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As I mentioned in the title, I want to find the expansion of $e^x$ using only the definition of $e=\lim_{n\to\infty} (1+1/n)^n$, and by developing it we find the sum of the infinite series. How do I prove it?

Bonnaduck
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procrastinating scholar
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1 Answers1

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We know that $e^x=\lim_{n\to\infty}(1+x/n)^n$. Expanding the binomial gives us:

$$\left(1+\frac xn\right)^n=\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}$$

Therefore,

\begin{align*} e^x&=\lim_{n\to\infty}\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}\\ &=\lim_{n\to\infty}\left[1+x+\frac{n(n-1)}{n^2}\cdot\frac{x^2}{2!}+\dots+\frac{n(n-1)\ldots(n-k+1)}{n^{n-k+1}}\frac{x^k}{k!}+\dots\right]\\ &=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\dots+\frac{x^k}{k!}+\dots\\ &=\sum_{k=0}^\infty\frac{x^k}{k!} \end{align*}

There are some details that would need to be filled in to make this a proper proof, but this is the general idea.

Bonnaduck
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  • this is the method that i went with , but there is an issue , the formula is correct to a certain k because we get 1 when n tends to infinity , but what about when k is equal to n the limit isn't 1 , using the same formula i get n!/(n^n) which tends to 0 – procrastinating scholar Nov 20 '21 at 19:19
  • @procrastinatingscholar, this argument is what I was thinking of as a "good heuristic", meaning that it quite correctly and strongly suggests the correct conclusion, but needs a little augmentation to be a proof. – paul garrett Nov 20 '21 at 19:27