Proof that $\exp(x)=\displaystyle\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}$

Theorem 1. $\exp(x)=\displaystyle\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}.$

Proof. Let $f(x)=\displaystyle\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}$ First, we’ll neglect the limits and expand $f(x)$ using the binomial theorem.

\[f_{n}(x)=\sum_{k=0}^{n}\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\]

Now, we will increase $n$ to a sufficiently large value such that $|f_{n}(x)-\exp(x)|<\epsilon,\forall x.$ Once we find such a value, our proof will be complete.

\[|\exp(x)-f_{n}(x)|=\left|\sum_{k=0}^{n}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)+\sum_{k=n+1}^{\infty}\frac{x^{k}}{k!}\right|.\]

We will make a slight change to the above equation and add a variable $m.$

\[|\exp(x)-f_{n}(x)|=\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)+\sum_{k=m+1}^{n}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)+\sum_{k=n+1}^{\infty}\frac{x^{k}}{k!}\right|\] \[\le\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|+\left|\sum_{k=m+1}^{n}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|+\left|\sum_{k=n+1}^{\infty}\frac{x^{k}}{k!}\right|\] \[\le\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|+\left|\sum_{k=m+1}^{n}\frac{x^{k}}{k!}\right|+\left|\sum_{k=n+1}^{\infty}\frac{x^{k}}{k!}\right|\] \[\le\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|+\left|\sum_{k=m+1}^{\infty}\frac{x^{k}}{k!}\right|\]

We will look at the 2 terms individually, starting with the right. Since $\displaystyle\frac{x^{k}}{k!}$ is eventually decreasing to 0, we can choose $m$ large enough so $\displaystyle\frac{x^{m}}{m!}<\displaystyle\frac{\epsilon}{4}$ and select an n larger than that.

\[\left|\sum_{k=m+1}^{\infty}\frac{x^{k}}{k!}\right|\le\frac{\epsilon}{4}\sum_{k=m+1}^{\infty}2^{-k}<\frac{\epsilon}{2}\]

Now, let’s look at the left sum. The following expression will help us understand why the left sum is small.

\[\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}=\frac{x^{k}(n(n-1)...(n-k+1)}{n^{k}k!}=\frac{x^{k}}{k!}\frac{(n(n-1)...(n-k+1)}{n^{k}}\approx\frac{x^{k}}{k!}\]

This approximation is valid as the terms almost cancel out. Because of this each of the terms will be fairly small so hopefully we can make the entire sum small. A clever observation is $n(n-m+1)\le (n-c)(n-k+1+c),$ $\forall k$ and positive $c$, this further means that

\[1\gt\frac{(n(n-1)\dots(n-k+1))}{n^k}\gt\left(\frac{n-m+1}{n}\right)^{m/2}\gt0\]

This further implies

\[\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|\le\left|\sum_{k=0}^{m}\frac{x^{k}}{k!}\left(1-\left(\frac{n-m+1}{n}\right)^{m/2}\right)\right|=\left|\left(1-\left(\frac{n-m+1}{n}\right)^{m/2}\right)\sum_{k=0}^{m}\frac{x^{k}}{k!}\right|\]

From this, we know that $\exp(x)$ is finite everywhere, $\exists c$ such that $c$ is greater than all the partial sums of $\exp(x)$, therefore

\[\left|\left(1-\left(\frac{n-m+1}{n}\right)^{m/2}\right)\sum_{k=0}^{m}\frac{x^{k}}{k!}\right|\lt c\left(1-\left(\frac{n-m+1}{n}\right)^{m/2}\right)\]

We can now use Bernoulli’s inequality to say that

Using this,

\[1>\left(\frac{n-m+1}{n}\right)^{m/2}=\left(1-\frac{m+1}{n}\right)^{m/2}>1-\frac{m^{2}+m}{2n}\] \[c\left(1-\left(\frac{n-m+1}{n}\right)^{m/2}\right)<\frac{c(m^{2}+m)}{2n}\]

We fixed m earlier, and we can use this to find out how large $n$ must be, we get

\[n>\frac{c(m^{2}+m)}{\epsilon}\]

Ergo, putting all of this together

\[\left|\exp(x)-f_{n}(x)\right|\le\left|\sum_{k=0}^{m}\left(\frac{x^{k}}{k!}-\binom{x}{n}^{k}\frac{n!}{k!(n-k)!}\right)\right|+\left|\sum_{k=m+1}^{\infty}\frac{x^{k}}{k!}\right|<\epsilon\]

This shows that $f_{n}\rightarrow \exp(x)$ pointwise, in other words, $\exp(x)=f(x)$.

$\blacksquare$

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