Exponential series – Serlo

The exponential series has the form $e=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}$ . We wil see that he limit $e$ 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 $((1+{\tfrac {1}{n}})^{n})_{n\in \mathbb {N} }$ and $((1+{\tfrac {1}{n}})^{n+1})_{n\in \mathbb {N} }$ . In this article, we will show the limit equivalence $\lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n}=\lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n+1}=e=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}$ which is far from obvious. Later, we will investigate a generalized form of the exponential series $\sum _{k=0}^{\infty }{\tfrac {x^{k}}{k!}}$ .

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

Proof (convergence of the exponential series)

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

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

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

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

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

Limit of the exponential series

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

That means, we need to show:

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

Proof (limit of the exponential series)

We show

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

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

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

Remarks

Alternatively, one may show $\liminf _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}\leq e\leq \limsup _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}$ , which also implies $\sum _{k=0}^{\infty }{\frac {1}{k!}}=e$ .
Further, the sequences $a_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}$ and $b_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}+{\tfrac {1}{n\cdot n!}}$ define nested intervals $(I_{n})_{n\in \mathbb {N} }=([a_{n},b_{n}])_{n\in \mathbb {N} }$ , where the real number included in all intervals is exactly $e$ .
The advantage if the exponential series compared to the sequences defining $e$ is that one can achieve much faster convergence. For instance, with 10 elements $\sum _{k=0}^{10}{\tfrac {1}{k!}}\approx 2.7182818011$ which is an approximation precise up to 7 digits: $e=2.718281828$ . By contrast, the 1000-th sequence element $\left(1+{\tfrac {1}{1000}}\right)^{1000}=2.7169239$ 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 $\sum _{k=0}^{\infty }{\tfrac {x^{k}}{k!}}$ . This series can be shown to converge for all $x\in \mathbb {R}$ . Therefore, it serves for a real-valued exponential function $\exp :\mathbb {R} \to \mathbb {R}$ . Even complex arguments are possible! However, $\exp(x)$ is a priori not the same as a power $e^{x}$ . So the computation rules for powers, like $\exp(x+y)=\exp(x)\exp(y)$ , must be shown explicitly.

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

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

Absolute convergence of a series →

Feedback? Do you want to join?

If you have questions concerning the content, or didn't understand something, the feel free to contact us! We would love to answer your questions! Also we are thankful for critics and/or comments! If you share our vision to explain university math in an comprehensible way, then contact us under:

E-Mail: en@serlo.org

This article is licensed under the free license CC-BY-SA 3.0. With that you can use it, modify it or share it freely, as long as you name „Serlo“ as source and put you changes under the same CC-BY-SA 3.0 oder an compatible license. On the page „Kopier uns!“ we explain you what you have to pay attention to, when using our texts, picture or videos.