# Overview: convergence criteria – Serlo

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Decision tree for convergence and divergence of series

We already introduced a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ as the sequence of the partial sums ${\displaystyle S_{n}=\sum _{k=1}^{n}a_{k}}$. A sequence is convergent, if the sequence of partial sums is convergent . Else the series is divergent. Assuming the series is convergent we define the value of the infinite sum of the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ to be equal to the limit of the sequence.

In this chapter we will study different criteria or tests to determine whether a series is convergent or not. In further chapters, we will study each of this criteria more attentively and give a proof for each.

## Criteria for convergence

We will give a proof for the following propositions in the respective main article for the criterion. Let a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be given. There is an arsenal of criteria to examine convergence:

### Absolute convergence

Main article: Absolute Konvergenz einer Reihe

Definition (Absolute convergence)

A series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is called absolutely convergent, if ${\displaystyle \sum _{k=1}^{\infty }|a_{k}|}$ is convergent.

Theorem (Absolute convergence)

If a series is absolutely convergent, it is also convergent. So if ${\displaystyle \sum _{k=1}^{\infty }|a_{k}|}$ is convergent, then ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is also convergent.

Example (Absolute convergence)

The series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}{\frac {1}{k^{2}}}}$ is convergent, because it is absolutely convergent. The series of absolute values ${\displaystyle \sum _{k=1}^{\infty }\left|(-1)^{k}{\frac {1}{k^{2}}}\right|=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ is convergent.

### Cauchy criterion

Main article: Cauchy-Kriterium für Reihen

Theorem (Cauchy criterion)

For all ${\displaystyle \varepsilon >0}$ let there be ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|<\varepsilon }$ for all ${\displaystyle n\geq m\geq N}$. Then the series is convergent.

Example (Cauchy criterion)

The geometric series ${\displaystyle \sum _{k=1}^{\infty }\left({\tfrac {1}{10}}\right)^{k}}$ is convergent according to the Cauchy criterion, because:

{\displaystyle {\begin{aligned}\left|\sum _{k=m}^{n}\left({\frac {1}{10}}\right)^{k}\right|&=\left|\left({\frac {1}{10}}\right)^{m}+\left({\frac {1}{10}}\right)^{m+1}+\left({\frac {1}{10}}\right)^{m+2}+\ldots +\left({\frac {1}{10}}\right)^{n}\right|\\[0.3em]&=\left|\left({\frac {1}{10}}\right)^{m}\right|\cdot \left|1+{\frac {1}{10}}+{\frac {1}{10^{2}}}+\ldots +{\frac {1}{10^{n-m}}}\right|\\[0.3em]&={\frac {1}{10^{m}}}\cdot \left|1+0{,}1+0{,}01+0{,}001+{\frac {1}{10^{n-m}}}\right|\\[0.3em]&={\frac {1}{10^{m}}}\cdot 1{,}\underbrace {111111\ldots 1} _{(n-m){\text{-mal}}}\\[0.3em]&\leq {\frac {1}{10^{m}}}\cdot 2\end{aligned}}}

Let ${\displaystyle \varepsilon >0}$. Since ${\displaystyle \displaystyle \lim _{m\to \infty }{\tfrac {2}{10^{m}}}=0}$ there is ${\displaystyle N\in \mathbb {N} }$ with ${\displaystyle \displaystyle {\tfrac {2}{10^{m}}}<\varepsilon }$ for all ${\displaystyle m\geq N}$. For this ${\displaystyle N}$ it follows from the above that ${\displaystyle \left|\sum _{k=m}^{n}\left({\tfrac {1}{10}}\right)^{k}\right|<\varepsilon }$ for all ${\displaystyle n\geq m\geq N}$. So we see that the series is convergent according to the Cauchy criterion.

### Leibniz criterion

Main article: Leibniz-Kriterium

Theorem (Leibniz criterion)

If the series has the form ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}b_{k}}$ and if the sequence ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ non-negative monotonic decreasing sequence null sequence , then the series is convergent.

Example (Leibniz criterion)

The convergence of the series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}{\frac {1}{2k-1}}}$ follows from the Leibniz criterion, because the sequence ${\displaystyle ({\tfrac {1}{2k-1}})_{k\in \mathbb {N} }}$ is a non-negative monotonic decreasing null sequence.

### Majorant criterion

Main article: Majorantenkriterium und Minorantenkriterium

Theorem (Majorant criterion)

Let ${\displaystyle |a_{k}|\leq b_{k}}$ for all ${\displaystyle k\in \mathbb {N} }$. If ${\displaystyle \sum _{k=1}^{\infty }b_{k}}$ is convergent, then the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is absolutely convergent.

Example (Majorant criterion)

We have ${\displaystyle {\tfrac {1}{2^{k}+k^{3}+7}}\leq {\tfrac {1}{2^{k}}}}$. Since the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}}}}$ is convergent (with limit 1), the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}+k^{3}+7}}}$ is also (absolutely) convergent.

### Ratio test

Main article: Quotientenkriterium

Theorem (Ratio test)

Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be a series with ${\displaystyle a_{k}\neq 0}$ for all ${\displaystyle k\in \mathbb {N} }$. If there exists a ${\displaystyle \theta <1}$ and a ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle \left|{\tfrac {a_{k+1}}{a_{k}}}\right|\leq \theta }$ for all ${\displaystyle k\geq N}$, then the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is absolutely convergent. This is particularly the case, if ${\displaystyle \lim _{k\to \infty }\left|{\tfrac {a_{k+1}}{a_{k}}}\right|<1}$ or ${\displaystyle \limsup _{k\to \infty }\left|{\tfrac {a_{k+1}}{a_{k}}}\right|<1}$.

Example (Ratio test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k!}}}$ is convergent, since we find that:

{\displaystyle {\begin{aligned}\lim _{k\to \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|&=\lim _{k\to \infty }\left|{\frac {\frac {1}{(k+1)!}}{\frac {1}{k!}}}\right|\\[0.5em]&=\lim _{k\to \infty }{\frac {k!}{(k+1)!}}\\[0.5em]&=\lim _{k\to \infty }{\frac {1}{k+1}}\\[0.5em]&=0<1\end{aligned}}}

### Root test

Main article: Wurzelkriterium

Theorem (Root test)

If ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1}$, then the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is absolutely convergent. In particular this is also true if ${\displaystyle \lim _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1}$.

Example (Root test)

The series ${\displaystyle \sum _{k=0}^{\infty }3^{-k}}$ is absolutely convergent, because we have:

${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=\limsup _{k\to \infty }{\sqrt[{k}]{|3^{-k}|}}=\limsup _{k\to \infty }3^{-1}=\lim _{k\to \infty }3^{-1}={\frac {1}{3}}<1}$

### Cauchy condensation test

Main article: Cauchysches Verdichtungskriterium

Theorem (Cauchy condensation test)

Let ${\displaystyle (a_{k})_{k\geq 1}}$ be a monotonically decreasing, real valued null sequence with ${\displaystyle a_{k}\geq 0}$ for all ${\displaystyle k\in \mathbb {N} }$. If ${\displaystyle \sum _{l=0}^{\infty }2^{l}a_{2^{l}}}$ is convergent, so is ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$).

Example (Cauchy condensation test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ is convergent according to Cauchy condensation test, because the series ${\displaystyle \sum _{l=0}^{\infty }2^{l}\left({\frac {1}{2^{l}}}\right)^{2}=\sum _{l=0}^{\infty }{\frac {1}{2^{l}}}=\sum _{l=0}^{\infty }\left({\frac {1}{2}}\right)^{l}}$ is convergent. Recall that ${\displaystyle \sum _{l=0}^{\infty }a^{l}}$ is convergent if ${\displaystyle |a|<1}$ (here we have ${\displaystyle a={\frac {1}{2}}<1}$).

### Integral test

Theorem (Integral test)

Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }f(k)}$, i.e. ${\displaystyle a_{k}=f(k)}$ for a function ${\displaystyle f:[1,\infty )\to \mathbb {R} }$. If ${\displaystyle f}$ is a monotonically decreasing function with non-negative values on the domain ${\displaystyle [1,\infty )}$ and if ${\displaystyle \int _{1}^{\infty }f(x)\,\mathrm {d} x<\infty }$, then the series is absolutely convergent.

Example (Integral test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ is absolutely convergent. We define ${\displaystyle f:[1,\infty )\to \mathbb {R} }$ with ${\displaystyle f(x)={\tfrac {1}{x^{2}}}}$. This function is a non-negative monotonically decreasing function, and now we can use the Integral test:

{\displaystyle {\begin{aligned}\int _{1}^{\infty }{\frac {1}{x^{2}}}\,\mathrm {d} x&=\lim _{a\to \infty }\int _{1}^{a}{\frac {1}{x^{2}}}\,\mathrm {d} x\\[0.5em]&=\lim _{a\to \infty }\left[-{\frac {1}{x}}\right]_{1}^{a}\\[0.5em]&=\lim _{a\to \infty }\left(-{\frac {1}{a}}+1\right)\\[0.5em]&=1<\infty \end{aligned}}}

Hint

We will give a proof that the Integral test works, after we have introduced Integrals. But for completeness purposes we listed it here. Please note that you can use this test only if it was proved in your lecture!

## Criteria for divergence

We are given a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$. To show that this series is divergent, there are multiple criteria:

### Term test

Main article: Trivialkriterium, Nullfolgenkriterium, Divergenzkriterium

Theorem (Term test)

If ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ diverges or ${\displaystyle \lim _{k\to \infty }a_{k}\neq 0}$, the the series is divergent.

Example (Term test)

The series ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {k+1}{k}}\right)^{k}}$ is divergent, because we have:

${\displaystyle \left({\frac {k+1}{k}}\right)^{k}=\left(1+{\frac {1}{k}}\right)^{k}\geq 1}$

Thus ${\displaystyle a_{k}=\left({\frac {k+1}{k}}\right)^{k}}$ cannot be a null sequence, which proves that ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {k+1}{k}}\right)^{k}}$ diverges.

### Cauchy Test

Main article: Cauchy-Kriterium für Reihen

Theorem (Cauchy-Kriterium)

If there is an ${\displaystyle \epsilon >0}$, so that for all ${\displaystyle N\in \mathbb {N} }$ there exist natural numbers ${\displaystyle {\tilde {n}}\geq {\tilde {m}}\geq N}$ with ${\displaystyle \left|\sum _{k={\tilde {m}}}^{\tilde {n}}a_{k}\right|\geq \epsilon }$, then the series is divergent.

Example (Cauchy Test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ is divergent according to the Cauchy test. Set ${\displaystyle \epsilon ={\tfrac {1}{4}}}$. For every ${\displaystyle N\in \mathbb {N} }$ we choose ${\displaystyle {\tilde {m}}=N+1}$ and ${\displaystyle {\tilde {n}}=2N}$. We then have:

{\displaystyle {\begin{aligned}\left|\sum _{k={\tilde {m}}}^{\tilde {n}}{\frac {1}{k}}\right|&=\left|\sum _{k=N+1}^{2N}{\frac {1}{k}}\right|\\[0.5em]&={\frac {1}{N+1}}+{\frac {1}{N+2}}+\ldots +{\frac {1}{2N}}\\[0.5em]&\geq {\frac {1}{2N}}+{\frac {1}{2N}}+\ldots +{\frac {1}{2N}}\\[0.5em]&=N\cdot {\frac {1}{2N}}\\[0.5em]&={\frac {1}{2}}\\&\geq \epsilon \end{aligned}}}

### Minorant criterion

Main article: Majorantenkriterium und Minorantenkriterium

Theorem (Minorant criterion)

Let ${\displaystyle a_{k}\geq c_{k}\geq 0}$ for almost all ${\displaystyle k\in \mathbb {N} }$. If ${\displaystyle \sum _{k=1}^{\infty }c_{k}}$ diverges, then also the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ diverges.

Example (Minorant criterion)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\sqrt {k}}}}$ is divergent, since we have ${\displaystyle {\tfrac {1}{\sqrt {k}}}\geq {\tfrac {1}{k}}}$ for all ${\displaystyle k\in \mathbb {N} }$, and the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ is divergent. In equations:

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\sqrt {k}}}\geq \sum _{k=1}^{\infty }{\frac {1}{k}}=\infty }$

### Quotient test

Main article: Quotientenkriterium

Theorem (Quotient test for divergence)

If ${\displaystyle \left|{\tfrac {a_{k+1}}{a_{k}}}\right|\geq 1}$ for almost all ${\displaystyle k\geq N}$ (i.e. for all ${\displaystyle k\geq K}$ for fixed ${\displaystyle K\in \mathbb {N} }$), then the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ diverges. In particular this is the case when${\displaystyle \lim _{k\to \infty }\left|{\tfrac {a_{k+1}}{a_{k}}}\right|>1}$.

Example (Quotient test for divergence)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {k!}{2^{k}}}}$ is divergent. Since we have:

{\displaystyle {\begin{aligned}\lim _{k\to \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|&=\lim _{k\to \infty }\left|{\frac {(k+1)!/2^{k+1}}{k!/2^{k}}}\right|\\[0.5em]&=\lim _{k\to \infty }{\frac {(k+1)!\cdot 2^{k}}{k!\cdot 2^{k+1}}}\\[0.5em]&=\lim _{k\to \infty }{\frac {k+1}{2}}\\[0.5em]&=\infty >1\end{aligned}}}

### Square root test

Main article: Wurzelkriterium

Theorem (Square root test)

If ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1}$, the the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is absolutely divergent. In particular, this is the case when ${\displaystyle \lim _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1}$.

Example (Square root test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {2^{k}}{k}}}$ diverges, because we have:

{\displaystyle {\begin{aligned}\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}&=\limsup _{k\to \infty }{\sqrt[{k}]{\left|{\frac {2^{k}}{k}}\right|}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {\sqrt[{k}]{2^{k}}}{\sqrt[{k}]{k}}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {2}{\sqrt[{k}]{k}}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ \lim _{k\to \infty }{\sqrt[{k}]{k}}=1\right.}\\[0.5em]&=2>1\end{aligned}}}

### Cauchy condensation test

Main article: Cauchysches Verdichtungskriterium

Theorem (Cauchy condensation test)

Let ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ be a monotonically decreasing real-valued null sequence with ${\displaystyle a_{k}\geq 0}$ for all ${\displaystyle k\in \mathbb {N} }$. If ${\displaystyle \sum _{l=0}^{\infty }2^{l}a_{2^{l}}}$ diverges, then also ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ diverges.

Example (Cauchy condensation test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ diverges, because ${\displaystyle {\frac {1}{k}}}$ is a monotonically decreasing null sequence and the series ${\displaystyle \sum _{l=0}^{\infty }2^{l}{\frac {1}{2^{l}}}=\sum _{l=0}^{\infty }1}$ diverges.

### Integral test

Theorem (Integral test)

Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }f(k)}$, so ${\displaystyle a_{k}=f(k)}$ for a function ${\displaystyle f:[1,\infty )\to \mathbb {R} }$. If ${\displaystyle f}$ on ${\displaystyle [1,\infty )}$ is a monotonically decreasing non-negative function, and if ${\displaystyle \int _{1}^{\infty }f(x)\,\mathrm {d} x=\infty }$, then the series is divergent.

Example (Integral test)

The series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ is divergent, because ${\displaystyle f:[1,\infty )\to \mathbb {R} }$ with ${\displaystyle f(x)={\tfrac {1}{x}}}$ is a monotonically decreasing non-negative function, and we have:

{\displaystyle {\begin{aligned}\int _{1}^{\infty }{\frac {1}{x}}\,\mathrm {d} x&=\lim _{a\to \infty }\int _{1}^{a}{\frac {1}{x}}\,\mathrm {d} x\\[0.5em]&=\lim _{a\to \infty }\left[\ln(x)\right]_{1}^{a}\\[0.5em]&=\lim _{a\to \infty }\left(\ln(a)-\ln(1)\right)\\&=\lim _{a\to \infty }\ln(a)\\&=\infty \end{aligned}}}

## Convergence is independent from starting index

In the section about the Cauchy test we saw that the starting index is irrelevant for the study of convergence. If we have a series of the form ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$, then we could also consider the series ${\displaystyle \sum _{k=10}^{\infty }a_{k}}$ or ${\displaystyle \sum _{k=4223}^{\infty }a_{k}}$. The only differences is the starting index ${\displaystyle k}$. This series all have the same convergence behaviour. So remember:

„If we remove or alter only finitely many members of the series, then the convergence behaviour doesn't change.“

If we remove or alter finitely many summands, the individual values of the series will change of course, but the convergence behaviour stays the same. This fact is useful, you should always keep it in the back of your head. This could be useful in those cases, where you are not interested in the exact values of the series, but only if it converges or not.

Example

Let ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ be defined as follows:

${\displaystyle a_{k}={\begin{cases}10^{k}&;k\leq 1000\\{\frac {1}{2^{k}}}&;k>1000\end{cases}}}$

Almost all members of the sequence ${\displaystyle \left(a_{k}\right)_{k\in \mathbb {N} }}$ are identical to ${\displaystyle \left({\tfrac {1}{2^{k}}}\right)_{k\in \mathbb {N} }}$ (only finitely many exceptions). Since the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}}}}$ is convergent, the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is also convergent, but the exact value of the limit is not the same.