# Exponential series – Serlo

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The exponential series has the form ${\displaystyle e=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}}$. We wil see that he limit ${\displaystyle e}$ indeed exists and is given by Euler's number, which we encountered first within the article Monotony criterion. There, it was defined as the limit of the seqeuences ${\displaystyle ((1+{\tfrac {1}{n}})^{n})_{n\in \mathbb {N} }}$ and ${\displaystyle ((1+{\tfrac {1}{n}})^{n+1})_{n\in \mathbb {N} }}$. In this article, we will show the limit equivalence ${\displaystyle \lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n}=\lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n+1}=e=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}}$ which is far from obvious. Later, we will investigate a generalized form of the exponential series ${\displaystyle \sum _{k=0}^{\infty }{\tfrac {x^{k}}{k!}}}$.

## Convergence of the exponential series

At first, we need to show that the exponential series converges at all:

Theorem (convergence of the exponential series)

The series ${\displaystyle \sum _{k=0}^{\infty }{\tfrac {1}{k!}}}$ converges.

Proof (convergence of the exponential series)

We need to prove that the partial sums ${\displaystyle (s_{n})_{n\in \mathbb {N} }=\left(\sum _{k=0}^{n}{\tfrac {1}{k!}}\right)_{n\in \mathbb {N} }}$ converges. This can be done using the monotony criterion for sequences, where the partial sums ${\displaystyle (s_{n})}$ form the sequence which we want to investigate.

Monotony is easy to see. Series elements are positive, so

${\displaystyle s_{n+1}=\sum _{k=0}^{n+1}{\frac {1}{k!}}=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{(n+1)!}} _{\geq 0}\geq \sum _{k=0}^{n}{\frac {1}{k!}}=s_{n}}$

Hence, ${\displaystyle (s_{n})}$ is monotonously increasing.

Boundedness from above can be shown by comparison to a geometric series with ${\displaystyle q={\tfrac {1}{2}}<1}$. For partial sums, we have

{\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {1}{k!}}&=\sum _{k=0}^{n}{\frac {1}{1\cdot 2\cdot 3\cdot \ldots \cdot (k-1)\cdot k}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{2\cdot 3\cdot \ldots \cdot (k-1)\cdot k}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\frac {1}{2}},{\frac {1}{3}},\ldots {\frac {1}{k-1}},{\frac {1}{k}}\leq {\frac {1}{2}}}\right.\\[0.5em]&\leq \sum _{k=0}^{n}\underbrace {\frac {1}{2\cdot 2\cdot \ldots \cdot 2\cdot 2}} _{(k-1){\text{ times}}}\\[0.5em]&=\sum _{k=0}^{n}\left({\frac {1}{2}}\right)^{k-1}\\[0.5em]&=2\cdot \sum _{k=0}^{n}\left({\frac {1}{2}}\right)^{k}\\[0.5em]&\leq 2\cdot \sum _{k=0}^{\infty }\left({\frac {1}{2}}\right)^{k}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{geometric series with }}q={\frac {1}{2}}}\right.\\[0.5em]&=2\cdot {\frac {1}{1-{\frac {1}{2}}}}\\[0.5em]&=2\cdot 2=4\end{aligned}}}

Hence, ${\displaystyle (s_{n})}$ is bounded from above by ${\displaystyle 4}$ . So the monotony criterion implies convergence.

## Limit of the exponential series

Now, we show that the exponential series indeed converges towards Euler's number ${\displaystyle e}$. This is done using the squeeze theorem by "squeezing" the partial sums ${\displaystyle (s_{n})_{n\in \mathbb {N} }=\left(\sum _{k=0}^{n}{\tfrac {1}{k!}}\right)_{n\in \mathbb {N} }}$ between the sequences ${\displaystyle ((1+{\tfrac {1}{n}})^{n})_{n\in \mathbb {N} }}$ and ${\displaystyle ((1+{\tfrac {1}{n}})^{n+1})_{n\in \mathbb {N} }}$ . Both bounding sequences converge to ${\displaystyle e}$ , so we get the desired result.

That means, we need to show:

${\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}}$

Theorem (limit of the exponential series)

There is ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}=e}$.

Proof (limit of the exponential series)

We show
${\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}}$
and use the squeeze theorem:

1st inequality: ${\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\tfrac {1}{k!}}}$. This is easier to establish than the second one. We need the binomial theorem ${\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}}$ with ${\displaystyle x={\tfrac {1}{n}}}$.

{\displaystyle {\begin{aligned}\left(1+{\frac {1}{n}}\right)^{n}&{\underset {\text{theorem}}{\overset {\text{binomial}}{=}}}\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot (n-k+1)}{k!}}\cdot {\frac {1}{n^{k}}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot (n-k+1)}{n\cdot n\cdot n\cdot \ldots \cdot n}}\cdot {\frac {1}{k!}}\\[0.5em]&=\sum _{k=0}^{n}\underbrace {{\frac {n}{n}}\cdot {\frac {n-1}{n}}\cdot {\frac {n-2}{n}}\cdot \ldots \cdot {\frac {n-k+1}{n}}} _{\leq 1}\cdot {\frac {1}{k!}}\\[0.5em]&\leq \sum _{k=0}^{n}{\frac {1}{k!}}\end{aligned}}}

2nd inequality: ${\displaystyle (1+{\tfrac {1}{n}})^{n+1}\geq \sum _{k=0}^{n}{\tfrac {1}{k!}}}$. Her, we additionally need the Bernoulli inequality ${\displaystyle (1+x)^{n}\geq 1+nx}$ for ${\displaystyle x\geq -1}$. In addition, a telescoping sum will appear in the end of the proof.

{\displaystyle {\begin{aligned}\left(1+{\frac {1}{n}}\right)^{n+1}&{\underset {\text{theorem}}{\overset {\text{binomial}}{=}}}\sum _{k=0}^{n+1}{\binom {n+1}{k}}{\frac {1}{n^{k}}}\\[0.5em]&=1+\sum _{k=1}^{n+1}{\binom {n+1}{k}}{\frac {1}{n^{k}}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{index shift}}}\right.\\[0.5em]&=1+\sum _{k=0}^{n}{\binom {n+1}{k+1}}{\frac {1}{n^{k+1}}}\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {(n+1)\cdot n\cdot (n-1)\cdot \ldots \cdot (n-k+1)}{(k+1)!}}\cdot {\frac {1}{n^{k+1}}}\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {(n+1)\cdot n\cdot (n-1)\cdot \ldots \cdot (n-(k-1))}{n\cdot n\cdot n\cdot \ldots \cdot n}}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}n+1,n,n-1,\ldots n-(k-1)\geq n-(k-1)}\right.\\[0.5em]&\geq 1+\sum _{k=0}^{n}{\frac {(n-(k-1))^{k+1}}{n^{k+1}}}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&=1+\sum _{k=0}^{n}\left(1-{\frac {k-1}{n}}\right)^{k+1}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Bernoulli's inequality with }}x=-{\frac {k-1}{n}}\geq -1}\right.\\[0.5em]&\geq 1+\sum _{k=0}^{n}\left(1-{\frac {(k+1)(k-1)}{n}}\right)\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Linearity of the sum}}}\right.\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {1}{(k+1)!}}-{\frac {1}{n}}\sum _{k=0}^{n}{\frac {(k+1)(k-1)}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Index shift (1st sum), cancellation (2nd sum)}}}\right.\\[0.5em]&=1+\sum _{k=1}^{n+1}{\frac {1}{k!}}-{\frac {1}{n}}\sum _{k=0}^{n}{\frac {(k-1)}{k!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}1={\frac {1}{1!}}{\text{ and factor out the }}n+1{\text{-th summand of the sum for }}k=0{\text{ and }}k=1{\text{ for the 2nd sum.}}}\right.\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{(n+1)!}} _{\geq 0}-{\frac {1}{n}}\left[-1+0+\sum _{k=2}^{n}{\frac {(k-1)}{k!}}\right]\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{re-formulate the fraction within the 2nd sum and use }}{\frac {k}{k!}}={\frac {1}{(k-1)!}}}\right.\\[0.5em]&\geq \sum _{k=0}^{n}{\frac {1}{k!}}-{\frac {1}{n}}\left[-1+\sum _{k=2}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)\right]\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{telescoping sum}}\sum _{k=2}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)=1-{\frac {1}{n!}}}\right.\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}-{\frac {1}{n}}\left[-1+1-{\frac {1}{n!}}\right]\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{n\cdot n!}} _{\geq 0}\\[0.5em]&\geq \sum _{k=0}^{n}{\frac {1}{k!}}\end{aligned}}}

In addition, we established ${\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}}$ . Since ${\displaystyle \lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n}=\lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n+1}=e}$, the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}=e}$.

## Remarks

• Alternatively, one may show ${\displaystyle \liminf _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}\leq e\leq \limsup _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}}$ , which also implies ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}=e}$.
• Further, the sequences ${\displaystyle a_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}}$ and ${\displaystyle b_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}+{\tfrac {1}{n\cdot n!}}}$ define nested intervals ${\displaystyle (I_{n})_{n\in \mathbb {N} }=([a_{n},b_{n}])_{n\in \mathbb {N} }}$, where the real number included in all intervals is exactly ${\displaystyle e}$.
• The advantage if the exponential series compared to the sequences defining ${\displaystyle e}$ is that one can achieve much faster convergence. For instance, with 10 elements ${\displaystyle \sum _{k=0}^{10}{\tfrac {1}{k!}}\approx 2.7182818011}$ which is an approximation precise up to 7 digits: ${\displaystyle e=2.718281828}$. By contrast, the 1000-th sequence element ${\displaystyle \left(1+{\tfrac {1}{1000}}\right)^{1000}=2.7169239}$ is precise to only 2 digits after the comma.

## Outlook: generalized exponential series

As remarked in the introduction, there is a generalization to the exponential series, which reads ${\displaystyle \sum _{k=0}^{\infty }{\tfrac {x^{k}}{k!}}}$ . This series can be shown to converge for all ${\displaystyle x\in \mathbb {R} }$ . Therefore, it serves for a real-valued exponential function ${\displaystyle \exp :\mathbb {R} \to \mathbb {R} }$ . Even complex arguments are possible! However, ${\displaystyle \exp(x)}$ is a priori not the same as a power ${\displaystyle e^{x}}$. So the computation rules for powers, like ${\displaystyle \exp(x+y)=\exp(x)\exp(y)}$, must be shown explicitly.

The series considered within this article is a special case with ${\displaystyle x=1}$:

${\displaystyle \exp(1)=\sum _{k=0}^{\infty }{\frac {1^{k}}{k!}}=\sum _{k=0}^{\infty }{\frac {1}{k!}}=e}$