How does one prove $(1+x/n)^n\to e^x$ pointwise through series expansion?

Question

We have $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} $$ and $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n \frac1{n^k}{n\choose k}x^k $$ therefore $$\left\lvert e^x-\left(1+\frac{x}{n}\right)^n\right\rvert\le\sum_{k=n+1}^\infty \frac{\lvert x\rvert^k}{k!}+\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert. $$

Question: As $n\to\infty$, the first sum trivially goes to $0$, but how to properly bound the second one?

I can show convergence in different ways but I'm interested in this specific approach.

Answer 1

\begin{align}\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert &\le\sum_{k=1}^{\infty}\frac{\lvert x\rvert^k}{k!} \underbrace{\left\lvert \big(\tfrac{n\cdot(n-1)\cdot\ldots\cdot(n-(k-1))}{n^k }- 1\big) \right\rvert}_{:=a_k(n)} \end{align}

Now let $\epsilon>0$ be arbitrary.

On the one hand $0\le a_k(n)\le 1$ for all $k$, hence the above series is bounded by $\exp(|x|)-1$. In particular there exists some $N$ such that $\sum_{k=N}^\infty a_k(n)\frac{|x|^k}{k!} \le \sum_{k=N}^\infty \frac{|x|^k}{k!} < \epsilon$.

On the other hand, since $a_k(n) \to 0$ for all $k$ we can then also find some $n_\epsilon$ such that $a_k(n_\epsilon) < \epsilon$ for all $k<N$. Putting the two together we conclude

\begin{align} \sum_{k=1}^{\infty}a_k(n_\epsilon)\frac{|x|^k}{k!} \le \sum_{k=1}^{N-1}\epsilon\frac{|x|^k}{k!} + \sum_{k=N}^{\infty}\frac{|x|^k}{k!} \le \epsilon(\exp(|x|)-1) + \epsilon = \exp(|x|)\epsilon \end{align}

Which demonstrates point-wise convergence.

Answer 2

All the coefficients in the 2nd sum are non-positive. Indeed,

\begin{multline*}\frac{1}{n^k}\binom{n}{k}-\frac{1}{k!}=\frac{1}{k!}\left(\frac{n!}{n^k(n-k)!}-1\right)=\frac{1}{k!}\left(\frac{(n-k+1)(n-k+2)\cdot\ldots\cdot n}{n^k}-1\right)\\\le \frac{1}{k!}\left(\frac{n^k}{n^k}-1\right) = 0.\end{multline*}

Performing your computation without absolute value and observing that for $x>0$ we have the difference on the left is positive, we have done for $x>0$. It seems to me that for $x<0$ the similar argument will work (with the reverted inequalities). Please check it. For $x=0$ the checking is direct.