# Exercises: Series – Serlo

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## Telescoping series

Exercise

Investigate, whether the following series converge. If so, compute their limits.

1. ${\displaystyle \sum _{k=1}^{\infty }\left({\sqrt {k+1}}-{\sqrt {k}}\right)}$
2. ${\displaystyle \sum _{k=1}^{\infty }{\frac {2k+1}{k^{2}(k+1)^{2}}}}$
3. ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k(k+2)}}}$
4. ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{4k^{2}+4k-3}}}$
5. ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}}$
6. ${\displaystyle \sum _{k=1}^{\infty }{\frac {k-1}{k(k+1)(k+2)}}}$

Hint regarding subtask 3: We have ${\displaystyle {\frac {1}{k(k+2)}}={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}}$. Why?

Hint regarding subtask 5: We have ${\displaystyle {\frac {1}{k(k+1)(k+2)}}={\frac {\frac {1}{2}}{k}}-{\frac {1}{k+1}}+{\frac {\frac {1}{2}}{k+2}}}$.

Solution

Subtask 1: This is a telescoping series with ${\displaystyle a_{k}={\sqrt {k}}}$. Taking a look at the partial sums we get

${\displaystyle \sum _{k=1}^{n}\left({\sqrt {k+1}}-{\sqrt {k}}\right)={\sqrt {n+1}}-{\sqrt {1}}}$

As ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ diverges, the series diverges as well.

Alternative solution: One can easily find a lower bound for the partial sum sequence:

{\displaystyle {\begin{aligned}s_{n}&=\sum \limits _{k=1}^{n}\left({\sqrt {k+1}}-{\sqrt {k}}\right)\\&=\sum \limits _{k=1}^{n}{\frac {\left({\sqrt {k+1}}-{\sqrt {k}}\right)\left({\sqrt {k+1}}+{\sqrt {k}}\right)}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&=\sum \limits _{k=1}^{n}{\frac {k+1-k}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&=\sum \limits _{k=1}^{n}{\frac {1}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&\geq \sum \limits _{k=1}^{n}{\frac {1}{{\sqrt {k+1}}+{\sqrt {k+1}}}}\\&=\sum \limits _{k=1}^{n}{\frac {1}{2{\sqrt {k+1}}}}\\&\geq \sum \limits _{k=1}^{n}{\frac {1}{2(k+1)}}\end{aligned}}}

As ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{2(k+1)}}=\infty }$ (harmonic series), ${\displaystyle (s_{n})}$ is not bounded from above/below. Hence, the series diverges.

{\displaystyle {\begin{aligned}{\frac {2k+1}{k^{2}(k+1)^{2}}}&={\frac {k^{2}+2k+1-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}}{k^{2}(k+1)^{2}}}-{\frac {k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {1}{k^{2}}}-{\frac {1}{(k+1)^{2}}}\end{aligned}}}

Obviously, this is a telescoping series with ${\displaystyle a_{k}={\tfrac {1}{k^{2}}}}$. We get:

${\displaystyle \sum _{k=1}^{\infty }{\frac {2k+1}{k^{2}(k+1)^{2}}}{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\sum _{k=1}^{\infty }\left({\frac {1}{k^{2}}}-{\frac {1}{(k+1)^{2}}}\right)=\lim _{n\to \infty }\left({\frac {1}{1^{2}}}-{\frac {1}{(n+1)^{2}}}\right)=1}$

Subtask 3: Take a look at the hint. We get

{\displaystyle {\begin{aligned}{\frac {1}{k(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\end{aligned}}}

This is a more generalized version of a telescoping sum. The first and last two summands do not cancel:

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{3}}\right)+\left({\frac {1}{2}}-{\frac {1}{4}}\right)+\left({\frac {1}{3}}-{\frac {1}{5}}\right)+\ldots +\left({\frac {1}{n-2}}-{\frac {1}{n}}\right)+\left({\frac {1}{n-1}}-{\frac {1}{n+1}}\right)+\left({\frac {1}{n}}-{\frac {1}{n+2}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{n+1}}-{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[1+{\frac {1}{2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {3}{2}}\\[0.5em]&={\frac {3}{4}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\frac {1}{4k^{2}+4k-3}}&={\frac {1}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot 4}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+4-2k)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+3-2k+1)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {2k+3}{(2k-1)(2k+3)}}-{\frac {2k-1}{(2k-1)(2k+3)}}\right)\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\end{aligned}}}

We get the following telescoping series:

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{4k^{2}+4k-3}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4k^{2}+4k-3}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\cdot \sum _{k=1}^{n}\left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\left[\left({\frac {1}{1}}-{\frac {1}{5}}\right)+\left({\frac {1}{3}}-{\frac {1}{7}}\right)+\left({\frac {1}{5}}-{\frac {1}{9}}\right)+\ldots +\left({\frac {1}{2n-5}}-{\frac {1}{2n-1}}\right)+\left({\frac {1}{2n-3}}-{\frac {1}{2n+1}}\right)+\left({\frac {1}{2n-1}}-{\frac {1}{2n+3}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{4}}\left[{\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{2n+1}}-{\frac {1}{2n+3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot \left[1+{\frac {1}{3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot {\frac {4}{3}}\\[0.5em]&={\frac {1}{3}}\end{aligned}}}

Subtask 5: Take a look at the hint! We get

{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {\frac {1}{2}}{k}}-{\frac {1}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\\[0.5em]\end{aligned}}}

Hence, we can calculate the series using two telescoping series:

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)-{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k+1}}-{\frac {1}{k+2}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{n+1}}\right)-\left({\frac {1}{2}}-{\frac {1}{n+2}}\right)\right]\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{2}}-{\frac {1}{n+1}}+{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0+0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}

Alternative solution: It holds that

{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+1)(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+1)(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\\[0.5em]\end{aligned}}}

Using the properties of telescoping series, we get:

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k(k+1)}}-{\frac {1}{(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1\cdot 2}}-{\frac {1}{(n+1)(n+2)}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}

Solution 6: It holds that

${\displaystyle {\frac {k-1}{k(k+1)(k+2)}}={\frac {k}{k(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}={\frac {1}{(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}}$

It follows that

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {k-1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {k-1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{shift}}{\overset {\text{Index}}{=}}}\lim \limits _{n\to \infty }\left[\sum _{k=2}^{n-1}{\frac {1}{k(k+1)}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&=\lim \limits _{n\to \infty }\left[\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&{\underset {\text{for limits}}{\overset {\text{Computation rules}}{=}}}\lim \limits _{n\to \infty }\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}} _{=1}-{\frac {1}{2}}-\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}} _{={\frac {1}{4}}{\text{ see }}4.}\\[0.5em]&=1-{\frac {1}{2}}-{\frac {1}{4}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}

## Geometric series

Exercise (Geometric series: Convergence and limits)

Investigate, whether the following series converge. If so, compute their limits.

1. ${\displaystyle \sum _{k=1}^{\infty }{\frac {3^{k}}{7^{k}}}}$
2. ${\displaystyle \sum _{k=0}^{\infty }\left(-{\frac {4}{3}}\right)^{k}}$
3. ${\displaystyle \sum _{k=0}^{\infty }{\frac {4}{(-3)^{k}}}}$
4. ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {2^{k}-1}{3^{k}}}\right)}$
5. ${\displaystyle \sum _{k=0}^{\infty }a_{k}}$ with ${\displaystyle a_{k}={\frac {1}{2^{k}}}}$ for ${\displaystyle k}$ being even and ${\displaystyle a_{k}={\frac {1}{3^{k}}}}$ for ${\displaystyle k}$ being odd.
6. ${\displaystyle \sum _{k=2}^{\infty }\left({\frac {3\cdot 2^{k+1}+9\cdot (-1)^{k}}{4\cdot 5^{k}}}\right)}$

Solution (Geometric series: Convergence and limits)

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {3^{k}}{7^{k}}}&=\sum _{k=0}^{\infty }\left({\frac {3}{7}}\right)^{k}-\underbrace {\frac {3^{0}}{7^{0}}} _{=1}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Geometric series with }}q={\frac {3}{7}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {3}{7}}}}-1\\[0.5em]&={\frac {1}{\frac {4}{7}}}-1\\[0.5em]&={\frac {7}{4}}-1\\[0.5em]&={\frac {3}{4}}\end{aligned}}}

Subtask 2: As ${\displaystyle |q|=\left|-{\tfrac {4}{3}}\right|={\tfrac {4}{3}}>1}$, this series diverges.

Subtask 3: The series ${\displaystyle \sum _{k=0}^{\infty }\left(-{\frac {1}{3}}\right)^{k}}$ converges. Using the computation rules for series, we get

{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {4}{(-3)^{k}}}&=\sum _{k=0}^{\infty }4\cdot \left(-{\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Factor rule for convergent series}}\right.}\\[0.5em]&=4\cdot \sum _{k=0}^{\infty }\left(-{\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Geometric series with }}q=-{\frac {1}{3}}\right.}\\[0.5em]&=4\cdot {\frac {1}{1-\left(-{\frac {1}{3}}\right)}}\\[0.5em]&=4\cdot {\frac {1}{\frac {4}{3}}}\\[0.5em]&=4\cdot {\frac {3}{4}}\\[0.5em]&=3\end{aligned}}}

Subtask 4: The series ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {2}{3}}\right)^{k}}$ and ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {1}{3}}\right)^{k}}$ converge. Using the computation rules for series, we get

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }\left({\frac {2^{k}-1}{3^{k}}}\right)&=\sum _{k=1}^{\infty }\left({\frac {2^{k}}{3^{k}}}-{\frac {1}{3^{k}}}\right)\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula for convergent series}}\right.}\\[0.5em]&=\sum _{k=1}^{\infty }\left({\frac {2}{3}}\right)^{k}-\sum _{k=1}^{\infty }\left({\frac {1}{3}}\right)^{k}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {2}{3}}\right)^{k}-\underbrace {\left({\frac {2}{3}}\right)^{0}} _{=1}-\sum _{k=0}^{\infty }\left({\frac {1}{3}}\right)^{k}+\underbrace {\left({\frac {1}{3}}\right)^{0}} _{=1}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {2}{3}}\right)^{k}-\sum _{k=0}^{\infty }\left({\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {2}{3}}}}-{\frac {1}{1-{\frac {1}{3}}}}\\[0.5em]&={\frac {1}{\frac {1}{3}}}-{\frac {1}{\frac {2}{3}}}\\[0.5em]&=3-{\frac {3}{2}}\\[0.5em]&={\frac {3}{2}}\end{aligned}}}

Subtask 5: The series ${\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{4}}\right)^{k}}$ and ${\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{9}}\right)^{k}}$ converge. Using the computation rule for series, we get

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k}&=\sum _{k=0}^{\infty }\left({\frac {1}{2^{2k}}}+{\frac {1}{3^{2k+1}}}\right)\\[0.5em]&=\sum _{k=0}^{\infty }\left(\left({\frac {1}{4}}\right)^{k}+{\frac {1}{3}}\left({\frac {1}{9}}\right)^{k}\right)\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula and factor rule for convergent series}}\right.}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {1}{4}}\right)^{k}+{\frac {1}{3}}\cdot \sum _{k=0}^{\infty }\left({\frac {1}{9}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {1}{4}}}}+{\frac {1}{3}}\cdot {\frac {1}{1-{\frac {1}{9}}}}\\[0.5em]&={\frac {1}{\frac {3}{4}}}+{\frac {1}{3}}\cdot {\frac {1}{\frac {8}{9}}}\\[0.5em]&={\frac {4}{3}}+{\frac {1}{3}}\cdot {\frac {9}{8}}\\[0.5em]&={\frac {4}{3}}+{\frac {3}{8}}\\[0.5em]&={\frac {32}{24}}+{\frac {9}{24}}\\[0.5em]&={\frac {41}{24}}\end{aligned}}}

Subtask 6: The series ${\displaystyle \sum _{k=0}^{\infty }\left({\frac {2}{5}}\right)^{k}}$ and ${\displaystyle \sum _{k=0}^{\infty }\left({\frac {1}{5}}\right)^{k}}$ converge. Using the computation rules for series, we get

{\displaystyle {\begin{aligned}\sum _{k=2}^{\infty }\left({\frac {3\cdot 2^{k+1}+9\cdot (-1)^{k}}{4\cdot 5^{k}}}\right)&=\sum _{k=0}^{\infty }\left({\frac {3\cdot 2\cdot 2^{k}+9\cdot (-1)^{k}}{4\cdot 5^{k}}}\right)-{\frac {3\cdot 2+9\cdot 1}{4\cdot 1}}-{\frac {3\cdot 4-9\cdot 1}{4\cdot 5}}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {6}{4}}\cdot {\frac {2^{k}}{5^{k}}}+{\frac {9}{4}}\cdot {\frac {(-1)^{k}}{5^{k}}}\right)-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula and factor rule for convergent series}}\right.}\\[0.5em]&={\frac {6}{4}}\cdot \sum _{k=0}^{\infty }\left({\frac {2}{5}}\right)^{k}+{\frac {9}{4}}\cdot \sum _{k=0}^{\infty }\left(-{\frac {1}{5}}\right)^{k}-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {3}{2}}\cdot {\frac {1}{1-{\frac {2}{5}}}}+{\frac {9}{4}}\cdot {\frac {1}{1+{\frac {1}{5}}}}-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&={\frac {3}{2}}\cdot {\frac {5}{3}}+{\frac {9}{4}}\cdot {\frac {5}{6}}-{\frac {75}{20}}-{\frac {3}{20}}\\[0.5em]&={\frac {5}{2}}+{\frac {15}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {20}{8}}+{\frac {15}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {35}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {175}{40}}-{\frac {156}{40}}\\[0.5em]&={\frac {19}{40}}\end{aligned}}}

## Harmonic series

Exercise (Harmonic series)

You may assume that ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ converges and that ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}}$ holds.

1. Explain why the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}}$, ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(2k)^{2}}}}$ and ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2}}}}$ converge.
2. Compute ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}}$ and ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2}}}}$.

Solution (Harmonic series)

1st series: The partial sum sequence ${\displaystyle (s_{n})=\left(\sum _{k=1}^{n}{\frac {1}{(2k-1)^{2}}}\right)}$ is monotonously increasing as all summands are positive. Futhermore, ${\displaystyle (s_{n})}$ is bounded from above as

${\displaystyle s_{n}=\sum _{k=1}^{n}{\frac {1}{(2k-1)^{2}}}\leq \sum _{k=1}^{n}{\frac {1}{k^{2}}}\leq \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}<\infty }$

Hence ${\displaystyle (s_{n})}$ converges.

2nd series: We know that ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ converges. Using the limit theorems for series, we get ${\displaystyle {\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=\sum _{k=1}^{\infty }{\frac {1}{4k^{2}}}=\sum _{k=1}^{\infty }{\frac {1}{(2k)^{2}}}}$. Hence, this series converges.

3rd series: As the series ${\displaystyle \sum _{k=1}^{\infty }\left|{\frac {(-1)^{k+1}}{k^{2}}}\right|=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}<\infty }$ converges absolutely, it converges.

1st series: We have

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}&=\sum _{k=1}^{\infty }\left({\frac {1}{(2k-1)^{2}}}+{\frac {1}{(2k)^{2}}}\right)\\[0.5em]&\left\downarrow \ {\text{both series converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}+\sum _{k=1}^{\infty }{\frac {1}{(2k)^{2}}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}+{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\end{aligned}}}

It follows that

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}&=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\cdot {\frac {\pi ^{2}}{6}}\\[0.5em]&={\frac {\pi ^{2}}{8}}\end{aligned}}}

2nd series: We have

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2}}}&=\sum _{k=1}^{\infty }\left({\frac {1}{(2k-1)^{2}}}-{\frac {1}{(2k)^{2}}}\right)\\[0.5em]&\left\downarrow \ {\text{both series converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {1}{2}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {1}{2}}\cdot {\frac {\pi ^{2}}{6}}\\[0.5em]&={\frac {\pi ^{2}}{12}}\end{aligned}}}

### Remark

Analogously, for the generalizes harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}}$ with ${\displaystyle \alpha >1}$ we can show:

• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{\alpha }}}=(1-{\frac {1}{2^{\alpha }}})\sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{\alpha }}}=(1-{\frac {1}{2^{\alpha -1}}})\sum _{k=1}^{\infty }{\frac {1}{k^{\alpha }}}}$

Exercise (Alternating harmonic series)

You may assume that ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}}$ converges and that ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}=\ln {2}}$ holds.

Explain why the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k(k+1)}}}$ converges and compute its limit.

Solution (Alternating harmonic series)

• Convergence: We will show that the series converges absolutely. In the article about absolute convergence we have proven that this implies convergence. Let ${\displaystyle (s_{n})=\left(\sum _{k=1}^{n}\left|{\frac {(-1)^{n+1}}{k(k+1)}}\right|\right)=\left(\sum _{k=1}^{n}{\frac {1}{k(k+1)}}\right)}$. As all summands are larger that zero, ${\displaystyle (s_{n})}$ increases monotonically. Furthermore, we have
${\displaystyle \sum _{k=1}^{n}{\frac {1}{k(k+1)}}{\overset {{\frac {1}{k+1}}\leq {\frac {1}{k}}}{\leq }}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\leq \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}<\infty }$.

Hence, ${\displaystyle (s_{n})}$ is bounded and converges.

• Limit: We have
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k(k+1)}}&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k(k+1)}}\\[0.5em]&\left\downarrow \ {\text{Partial fraction decomposition}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)\\[0.5em]&\left\downarrow \ {\text{series }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}{\text{ and }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k+1}}{\text{ converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}-\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k+1}}\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=1}^{\infty }(-1)^{k+2}{\frac {1}{k+1}}\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=0}^{\infty }(-1)^{k+2}{\frac {1}{k+1}}-1\\[0.5em]&\left\downarrow \ {\text{Index shift (2nd series)}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}-1\\[0.5em]&=\ln(2)+\ln(2)-1\\[0.5em]&=2\ln(2)-1\end{aligned}}}

## e-series

Exercise (e-series)

Explain why the following series converge and compute their limits:

1. ${\displaystyle \sum _{k=0}^{\infty }{\frac {k-1}{k!}}=-1+0+{\frac {1}{2!}}+{\frac {2}{3!}}+{\frac {3}{4!}}+\ldots }$
2. ${\displaystyle \sum _{k=0}^{\infty }{\frac {k+1}{k!}}=1+{\frac {2}{1!}}+{\frac {3}{2!}}+{\frac {4}{3!}}+\ldots }$

Solution (e-series)

Subtask 1: The partial sum sequence ${\displaystyle (s_{n})=\left(\sum _{k=0}^{n}{\frac {k-1}{k!}}\right)}$ increases monotonously and is bounded from above as

{\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {k-1}{k!}}&\leq \sum _{k=0}^{n}{\frac {k}{k!}}\\[0.5em]&=\sum _{k=1}^{n}{\frac {k}{k!}}\\[0.5em]&=\sum _{k=1}^{n}{\frac {1}{(k-1)!}}\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{n-1}{\frac {1}{k!}}\\[0.5em]&\leq \sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=e<\infty \end{aligned}}}

Hence, the sequence ${\displaystyle (s_{n})}$ converges.

Furthermore, we have

{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k-1}{k!}}&=\sum _{k=0}^{\infty }\left({\frac {k}{k!}}-{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{series converge}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {k}{k!}}-\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {k}{k!}}-e\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(k-1)!}}-e\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {1}{k!}}-e\\[0.5em]&=e-e\\[0.5em]&=0\end{aligned}}}

Alternative solution: Via telescoping sum. We have

{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k-1}{k!}}&=-1+\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {k}{k!}}-{\frac {1}{k!}}\right)\\[0.5em]&=-1+\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{Telescoping sum}}\right.\\[0.5em]&=-1+\lim _{n\to \infty }\left({\frac {1}{0!}}-{\frac {1}{n!}}\right)\\[0.5em]&=-1+\left(1-\lim _{n\to \infty }{\frac {1}{n!}}\right)\\[0.5em]&=-1+1-0\\[0.5em]&=0\end{aligned}}}

Subtask 2: The partial sum sequence ${\displaystyle (s_{n})=\left(\sum _{k=0}^{n}{\frac {k+1}{k!}}\right)}$ increases monotonously and is bounded from above as

{\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {k+1}{k!}}&\leq \sum _{k=0}^{n}{\frac {k+k}{k!}}\\[0.5em]&=2\sum _{k=1}^{n}{\frac {k}{k!}}\\[0.5em]&=2\sum _{k=1}^{n}{\frac {1}{(k-1)!}}\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=2\sum _{k=0}^{n-1}{\frac {1}{k!}}\\[0.5em]&\leq 2\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=2e<\infty \end{aligned}}}

Hence, the sequence ${\displaystyle (s_{n})}$ converges.

Furthermore, we have

{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k+1}{k!}}&=\sum _{k=0}^{\infty }\left({\frac {k}{k!}}+{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{series converge}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {k}{k!}}+\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {k}{k!}}+e\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(k-1)!}}+e\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {1}{k!}}+e\\[0.5em]&=e+e\\[0.5em]&=2e\end{aligned}}}

## Rearrangement theorem for series

Exercise (Rearrangement of alternating harmonic series)

The alternating harmonic series

${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+{\frac {1}{7}}-{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+\ldots +{\frac {1}{2n-1}}-{\frac {1}{2n}}\pm \ldots }$

and

${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k^{2}}}=1-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+{\frac {1}{5^{2}}}-{\frac {1}{6^{2}}}+{\frac {1}{7^{2}}}-{\frac {1}{8^{2}}}+{\frac {1}{9^{2}}}-{\frac {1}{10^{2}}}+\ldots +{\frac {1}{(2n-1)^{2}}}-{\frac {1}{(2n)^{2}}}\pm \ldots }$

converge to ${\displaystyle S}$ resp. ${\displaystyle T}$. Show that the following rearrangements converge to the limits given.

1. ${\displaystyle {\color {green}1+{\frac {1}{3}}}{\color {Red}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {green}+{\frac {1}{5}}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{6}}-{\frac {1}{8}}}+\ldots {\color {green}+{\frac {1}{4n-3}}+{\frac {1}{4n-1}}}{\color {Red}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\pm \ldots =S}$
2. ${\displaystyle {\color {green}1+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {green}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{15}}}{\color {Red}-{\frac {1}{6}}-{\frac {1}{8}}}+\ldots {\color {green}+{\frac {1}{8n-7}}+{\frac {1}{8n-5}}+{\frac {1}{8n-3}}+{\frac {1}{8n-1}}}{\color {Red}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\pm \ldots ={\frac {3}{2}}S}$
3. ${\displaystyle {\color {green}1+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}}{\color {Red}-{\frac {1}{2^{2}}}-{\frac {1}{4^{2}}}}{\color {green}+{\frac {1}{9^{2}}}+{\frac {1}{11^{2}}}+{\frac {1}{13^{2}}}+{\frac {1}{15^{2}}}}{\color {Red}-{\frac {1}{6^{2}}}-{\frac {1}{8^{2}}}}+\ldots {\color {green}+{\frac {1}{(8n-7)^{2}}}+{\frac {1}{(8n-5)^{2}}}+{\frac {1}{(8n-3)^{2}}}+{\frac {1}{(8n-1)^{2}}}}{\color {Red}-{\frac {1}{(4n-2)^{2}}}-{\frac {1}{(4n)^{2}}}}\pm \ldots =T}$

Hint regarding subtask 2: Start off showing that ${\displaystyle S_{8n}+{\frac {1}{2}}S_{4n}=T_{6n}}$ with ${\displaystyle S_{n}}$ being the ${\displaystyle n}$-th partial sum of the alternating harmonic series and ${\displaystyle T_{n}}$ being the ${\displaystyle n}$-th partial sum of the rearranges series.

Solution (Rearrangement of alternating harmonic series)

Subtask 1: With ${\displaystyle S_{n}=\sum _{k=1}^{n}{\frac {(-1)^{k+1}}{k}}}$ and ${\displaystyle T_{m}=\sum _{k=1}^{m}{\frac {(-1)^{\sigma (k+1)}}{\sigma (k)}}}$ being the partial sums of the alternating harmonic series and the first rearrangement, we have:

{\displaystyle {\begin{aligned}T_{4n}&={\color {green}1+{\frac {1}{3}}}{\color {Red}-{\frac {1}{2}}-{\frac {1}{4}}}{\color {green}+{\frac {1}{5}}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{6}}-{\frac {1}{8}}}{\color {green}+}\ldots {\color {green}+{\frac {1}{4n-3}}+{\frac {1}{4n-1}}}{\color {Red}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\\&=\left({\color {green}1+{\frac {1}{3}}}{\color {Red}-{\frac {1}{2}}-{\frac {1}{4}}}\right)+\left({\color {green}{\frac {1}{5}}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{6}}-{\frac {1}{8}}}\right)+\ldots +\left({\color {green}{\frac {1}{4n-3}}+{\frac {1}{4n-1}}}{\color {Red}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\right)\\&=\left({\color {green}1}{\color {Red}-{\frac {1}{2}}}{\color {green}+{\frac {1}{3}}}{\color {Red}-{\frac {1}{4}}}\right)+\left({\color {green}{\frac {1}{5}}}{\color {Red}-{\frac {1}{6}}}{\color {green}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{8}}}\right)+\ldots +\left({\color {green}{\frac {1}{4n-3}}}{\color {Red}-{\frac {1}{4n-2}}}{\color {green}+{\frac {1}{4n-1}}}{\color {Red}-{\frac {1}{4n}}}\right)\\&={\color {green}1}{\color {Red}-{\frac {1}{2}}}{\color {green}+{\frac {1}{3}}}{\color {Red}-{\frac {1}{4}}}{\color {green}+{\frac {1}{5}}}{\color {Red}-{\frac {1}{6}}}{\color {green}+{\frac {1}{7}}}{\color {Red}-{\frac {1}{8}}}{\color {green}+}\ldots {\color {green}+{\frac {1}{4n-3}}}{\color {Red}-{\frac {1}{4n-2}}}{\color {green}+{\frac {1}{4n-1}}}{\color {Red}-{\frac {1}{4n}}}\\&=S_{4n}\end{aligned}}}

${\displaystyle (S_{4n})_{n\in \mathbb {N} }}$ converges and hence,${\displaystyle (S_{n})_{n\in \mathbb {N} }}$ tends to ${\displaystyle S}$. Thus, ${\displaystyle (T_{4n})_{n\in \mathbb {N} }}$ converges and ${\displaystyle (T_{n})_{n\in \mathbb {N} }}$ tends to ${\displaystyle S}$ as well.

{\displaystyle {\begin{aligned}S_{8n}+{\frac {1}{2}}S_{4n}&=\sum _{k=1}^{8n}(-1)^{k+1}{\frac {1}{k}}+{\frac {1}{2}}\left(\sum _{k=1}^{4n}(-1)^{k+1}{\frac {1}{k}}\right)\\&=\left(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+{\frac {1}{7}}-{\frac {1}{8}}\right)+\ldots +\left({\frac {1}{8n-7}}-{\frac {1}{8n-6}}+{\frac {1}{8n-5}}-{\frac {1}{8n-4}}+{\frac {1}{8n-3}}-{\frac {1}{8n-2}}+{\frac {1}{8n-1}}-{\frac {1}{8n}}\right)\\&\quad +{\frac {1}{2}}\left(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}\right)+\ldots +{\frac {1}{2}}\left({\frac {1}{4n-3}}-{\frac {1}{4n-2}}+{\frac {1}{4n-1}}-{\frac {1}{4n}}\right)\\&=\left(1{\color {Orange}-{\frac {1}{2}}}+{\frac {1}{3}}{\color {blue}-{\frac {1}{4}}}+{\frac {1}{5}}{\color {Orange}-{\frac {1}{6}}}+{\frac {1}{7}}{\color {blue}-{\frac {1}{8}}}\right)+\ldots +\left({\frac {1}{8n-7}}{\color {Orange}-{\frac {1}{8n-6}}}+{\frac {1}{8n-5}}{\color {blue}-{\frac {1}{8n-4}}}+{\frac {1}{8n-3}}{\color {Orange}-{\frac {1}{8n-2}}}+{\frac {1}{8n-1}}{\color {blue}-{\frac {1}{8n}}}\right)\\&\quad +\left({\color {Orange}{\frac {1}{2}}}{\color {blue}-{\frac {1}{4}}}{\color {Orange}+{\frac {1}{6}}}{\color {blue}-{\frac {1}{8}}}\right)+\ldots +\left({\color {Orange}{\frac {1}{8n-6}}}{\color {blue}-{\frac {1}{8n-4}}}{\color {Orange}+{\frac {1}{8n-2}}}{\color {blue}-{\frac {1}{8n}}}\right)\\&=\left(1+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}{\color {blue}-{\frac {2}{4}}-{\frac {2}{8}}}\right)+\ldots +\left({\frac {1}{8n-7}}+{\frac {1}{8n-5}}+{\frac {1}{8n-3}}+{\frac {1}{8n-1}}{\color {blue}-{\frac {2}{8n-4}}-{\frac {2}{8n}}}\right)\\&=\left(1+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}{\color {blue}-{\frac {1}{2}}-{\frac {1}{4}}}\right)+\ldots +\left({\frac {1}{8n-7}}+{\frac {1}{8n-5}}+{\frac {1}{8n-3}}+{\frac {1}{8n-1}}{\color {blue}-{\frac {1}{4n-2}}-{\frac {1}{4n}}}\right)\\&=T_{6n}\end{aligned}}}
As ${\displaystyle (S_{8n})}$ and ${\displaystyle (S_{4n})}$ tend to ${\displaystyle S}$, ${\displaystyle S_{8n}+{\frac {1}{2}}S_{4n}}$ tends to ${\displaystyle S+{\frac {1}{2}}S={\frac {3}{2}}S}$. We can conclude that ${\displaystyle (T_{6n})}$ converges and ${\displaystyle (T_{n})}$ tends to ${\displaystyle {\frac {3}{2}}S}$.
Subtask 3: As ${\displaystyle \sum _{k=1}^{\infty }\left|{\frac {(-1)^{k+1}}{k^{2}}}\right|=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}<\infty }$, the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2}}}}$ converges absolutely. Using the rearrangement theorem for absolutely convergent series, we get that every rearrangement of the series converges and tends to the same limit. Hence, the given rearrangement tends to ${\displaystyle T}$.
Prove the following statements: Let ${\displaystyle \sum \limits _{k=1}^{\infty }a_{k}}$ be a convergent -but not absolutely convergent - series. Then there exists a rearrangement of the series that...
1. diverges but not to ${\displaystyle \infty }$ or ${\displaystyle -\infty }$.
2. converges to an arbitrary ${\displaystyle S\in \mathbb {R} }$.