Exponential series – "Math for Non-Geeks"

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

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

Proof (convergence of the exponential series)

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

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

$s_{n+1}=\sum _{k=0}^{n+1}\frac{1}{k!}=\sum _{k=0}^n\frac{1}{k!}+\underset{\ge 0}{\underbrace{\frac{1}{(n+1)!}}}\ge \sum _{k=0}^n\frac{1}{k!}=s_n$

Hence, $(s_n)$ is monotonically increasing.

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

$\begin{array}{cc}\sum _{k=0}^n\frac{1}{k!}& =\sum _{k=0}^n\frac{1}{1\cdot 2\cdot 3\cdot \dots \cdot (k-1)\cdot k}\\ & =\sum _{k=0}^n\frac{1}{2\cdot 3\cdot \dots \cdot (k-1)\cdot k}\\ & \downarrow \frac{1}{2},\frac{1}{3},\dots \frac{1}{k-1},\frac{1}{k}\le \frac{1}{2}\\ & \le \sum _{k=0}^n\underset{(k-1)\text{ times}}{\underbrace{\frac{1}{2\cdot 2\cdot \dots \cdot 2\cdot 2}}}\\ & =\sum _{k=0}^n{\left(\frac{1}{2}\right)}^{k-1}\\ & =2\cdot \sum _{k=0}^n{\left(\frac{1}{2}\right)}^k\\ & \le 2\cdot \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^k\\ & \downarrow \text{geometric series with }q=\frac{1}{2}\\ & =2\cdot \frac{1}{1-\frac{1}{2}}\\ & =2\cdot 2=4\end{array}$

Hence, $(s_n)$ is bounded from above by $4$ . So the monotonicity 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 $(s_n{)}_{n\in \mathbb{N}}={\left(\sum _{k=0}^n\frac{1}{k!}\right)}_{n\in \mathbb{N}}$ between the sequences $((1+\frac{1}{n}{)}^n{)}_{n\in \mathbb{N}}$ and $((1+\frac{1}{n}{)}^{n+1}{)}_{n\in \mathbb{N}}$ . Both bounding sequences converge to $e$ , so we get the desired result.

That means, we need to show: $(1+\frac{1}{n}{)}^n\le \sum _{k=0}^n\frac{1}{k!}\le (1+\frac{1}{n}{)}^{n+1}$

Theorem (limit of the exponential series)

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

Proof (limit of the exponential series)

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

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

$\begin{array}{cc}{(1+\frac{1}{n})}^n& \underset{\text{theorem}}{\stackrel{\text{binomial}}{=}}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{1}{n^k}\\ & =\sum _{k=0}^n\frac{n\cdot (n-1)\cdot (n-2)\cdot \dots \cdot (n-k+1)}{k!}\cdot \frac{1}{n^k}\\ & =\sum _{k=0}^n\frac{n\cdot (n-1)\cdot (n-2)\cdot \dots \cdot (n-k+1)}{n\cdot n\cdot n\cdot \dots \cdot n}\cdot \frac{1}{k!}\\ & =\sum _{k=0}^n\underset{\le 1}{\underbrace{\frac{n}{n}\cdot \frac{n-1}{n}\cdot \frac{n-2}{n}\cdot \dots \cdot \frac{n-k+1}{n}}}\cdot \frac{1}{k!}\\ & \le \sum _{k=0}^n\frac{1}{k!}\end{array}$

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

$\begin{array}{cc}{(1+\frac{1}{n})}^{n+1}& \underset{\text{theorem}}{\stackrel{\text{binomial}}{=}}\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n+1}{k})\frac{1}{n^k}\\ & =1+\sum _{k=1}^{n+1}(\genfrac{}{}{0ex}{}{n+1}{k})\frac{1}{n^k}\\ & \downarrow \text{index shift}\\ & =1+\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n+1}{k+1})\frac{1}{n^{k+1}}\\ & =1+\sum _{k=0}^n\frac{(n+1)\cdot n\cdot (n-1)\cdot \dots \cdot (n-k+1)}{(k+1)!}\cdot \frac{1}{n^{k+1}}\\ & =1+\sum _{k=0}^n\frac{(n+1)\cdot n\cdot (n-1)\cdot \dots \cdot (n-(k-1))}{n\cdot n\cdot n\cdot \dots \cdot n}\cdot \frac{1}{(k+1)!}\\ & \downarrow n+1,n,n-1,\dots n-(k-1)\ge n-(k-1)\\ & \ge 1+\sum _{k=0}^n\frac{(n-(k-1){)}^{k+1}}{n^{k+1}}\cdot \frac{1}{(k+1)!}\\ & =1+\sum _{k=0}^n{(1-\frac{k-1}{n})}^{k+1}\cdot \frac{1}{(k+1)!}\\ & \downarrow \text{Bernoulli's inequality with }x=-\frac{k-1}{n}\ge -1\\ & \ge 1+\sum _{k=0}^n(1-\frac{(k+1)(k-1)}{n})\cdot \frac{1}{(k+1)!}\\ & \downarrow \text{Linearity of the sum}\\ & =1+\sum _{k=0}^n\frac{1}{(k+1)!}-\frac{1}{n}\sum _{k=0}^n\frac{(k+1)(k-1)}{(k+1)!}\\ & \downarrow \text{Index shift (1st sum), cancellation (2nd sum)}\\ & =1+\sum _{k=1}^{n+1}\frac{1}{k!}-\frac{1}{n}\sum _{k=0}^n\frac{(k-1)}{k!}\\ & \downarrow 1=\frac{1}{1!}\text{ and factor out the }n+1\text{-th summand of the sum for }k=0\text{ and }k=1\text{ for the 2nd sum.}\\ & =\sum _{k=0}^n\frac{1}{k!}+\underset{\ge 0}{\underbrace{\frac{1}{(n+1)!}}}-\frac{1}{n}[-1+0+\sum _{k=2}^n\frac{(k-1)}{k!}]\\ & \downarrow \text{re-formulate the fraction within the 2nd sum and use }\frac{k}{k!}=\frac{1}{(k-1)!}\\ & \ge \sum _{k=0}^n\frac{1}{k!}-\frac{1}{n}[-1+\sum _{k=2}^n(\frac{1}{(k-1)!}-\frac{1}{k!})]\\ & \downarrow \text{telescoping sum}\sum _{k=2}^n(\frac{1}{(k-1)!}-\frac{1}{k!})=1-\frac{1}{n!}\\ & =\sum _{k=0}^n\frac{1}{k!}-\frac{1}{n}[-1+1-\frac{1}{n!}]\\ & =\sum _{k=0}^n\frac{1}{k!}+\underset{\ge 0}{\underbrace{\frac{1}{n\cdot n!}}}\\ & \ge \sum _{k=0}^n\frac{1}{k!}\end{array}$

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

Remarks

• Alternatively, one may show $\underset{n\to \mathrm{\infty }}{lim\; inf}\sum _{k=0}^n\frac{1}{k!}\le e\le \underset{n\to \mathrm{\infty }}{lim\; sup}\sum _{k=0}^n\frac{1}{k!}$ , which also implies $\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}=e$.
• Further, the sequences $a_n=\sum _{k=0}^n\frac{1}{k!}$ and $b_n=\sum _{k=0}^n\frac{1}{k!}+\frac{1}{n\cdot n!}$ 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 $\sum _{k=0}^{10}\frac{1}{k!}\approx 2.7182818011$ which is an approximation precise up to 7 digits: $e=2.718281828$. By contrast, the 1000-th sequence element ${(1+\frac{1}{1000})}^{1000}=2.7169239$ 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 $\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}$ . This series can be shown to converge for all $x\in ℝ$ . Therefore, it serves for a real-valued exponential function $\exp :ℝ\to ℝ$ . 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 $x=1$:

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