- #1

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## Homework Statement

[itex]e^{x}=\sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!}[/itex] and [itex]e^{x}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}[/itex].

I want to show that [itex]\sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}[/itex].

**2. The attempt at a solution**

Let $$t_{n}=\left(1+\frac{x}{n}\right)^{n}.$$ Then by the binomial theorem we have

$$ t_{n}=1+x+\frac{x^{2}}{2!}\left(\frac{n-1}{n}\right)+\frac{x^{3}}{3!}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)+\cdots+\frac{x^{n}}{n!}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{1}{n}\right)=\sum\limits_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}.$$

Then

$$

\begin{align*}

\lim\limits_{n\rightarrow\infty}t_{n}=\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}

&=\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty}\frac{x^{k}}{k!}

\frac{n!}{n^{k}(n-k)!}\\

&=1+x+\frac{x^{2}}{2!}\lim\limits_{n\rightarrow\infty}\frac{n-1}{n}+\frac{x^{3}}{3!}\lim\limits_{n\rightarrow\infty}\frac{n-1}{n}\frac{n-2}{n}+\cdots \\

&=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots \\

&=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}.

\end{align*}

$$

Hence

$$\lim\limits_{n\rightarrow\infty}t_{n}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=\sum\limits_{k=0}^{\infty}\frac{x^{k}}{k!}.$$

Can someone check if my reasoning works? I am particularly worried about the part where I say

$$

\lim\limits_{n\rightarrow\infty} \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{n!}{n^{k}(n-k)!}=\sum_{k=0}^{\infty}\lim\limits_{n\rightarrow\infty}\frac{x^{k}}{k!}

\frac{n!}{n^{k}(n-k)!}.

$$