Math for Non-Geeks/Exponential series

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The exponential series has the form $e=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}$. We wil see that he limit $e$ indeed exists and is given by Euler's number, which we encountered first within the article Monotonicity criterion. There, it was defined as the limit of the seqeuences $((1+\frac{1}{n}{)}^n{)}_{n\in \mathbb{N}}$ 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:

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: Template:Math

Remarks

• Alternatively, one may show $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; inf}}\sum _{k=0}^n\frac{1}{k!}\le e\le \underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sum _{k=0}^n\frac{1}{k!}$ , 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 $x=1$:

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