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We 'know' that $$\lim_{n\to\infty}\left(1+\frac 1n\right)^n=\sum_{n=0}^\infty\frac 1{n!}$$ To prove this, I have started with the inequality $$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac1{k!}\prod_{j=0}^{k-1}\frac{n-j}n\le\sum_{k=0}^n\frac1{k!}$$ Then I tried to find a strictly increasing sequence of natural numbers $a_n$ such that the following inequality holds for every $n$: $$\left(1+\frac1{a_n}\right)^{a_n}\ge\sum_{k=0}^n\frac 1{k!}$$ I have tried with $a_n=n+1$ and $a_n=2n$, but I couldn't prove it. In fact, I tried some terms with a calculator and it seems to be false for these particular sequences of natural numbers. Is this the way to prove the equality? If it is not, how?

EDIT: I should have said this before, but exp and log functions are not already defined. So the answers to the 'duplicate' are useless to me.

ajotatxe
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    Just to give a suggestion along the lines of the method you had in mind (since the answers in the "duplicate" follow a rather different methodology), you have $(1+1/n)^n=\sum_{k=0}^n \frac{1}{k!} \frac{n!}{n^k (n-k)!}$ and so what you are left to do is an interchange of limit and summation, since $\lim_{n \to \infty} \frac{n!}{n^k (n-k)!}=1$ for each fixed $k$. One way to justify this interchange is to exploit the fact that $ \frac{n!}{n^k (n-k)!}$ for each fixed $k$ is actually an increasing function of $n$. – Ian May 28 '21 at 19:48
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    I have done it herewith dominated convergence --> https://math.stackexchange.com/a/2172435/399263, but there are probably many duplicates out there. As stated by Ian the starting point is the binomial formula. You can also find a proof with your method here --> https://math.stackexchange.com/a/1898375/399263 in Santiago's answer. – zwim May 28 '21 at 20:49
  • Have a look at https://math.stackexchange.com/a/831803/42969, where the equality is shown without reference to exp or log functions. – Martin R May 29 '21 at 09:25
  • This has been discussed many times over. See for example https://math.stackexchange.com/a/901708/72031 – Paramanand Singh May 29 '21 at 11:46

1 Answers1

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It is easy to show that both $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ and $\sum_{k=0}^\infty \frac1{k!}$ exist. We shall use this in that which follows.

First, you were correct to begin by using the binomial theorem to write

$$\begin{align} \left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\ &=\sum_{k=0}^n \frac1{k!}\prod_{j=0}^{k-1} \left(1-\frac{j}{n}\right)\\\\ &\le \sum_{k=0}^n \frac1{k!}\tag1 \end{align}$$

Hence, we see that from $(1)$ that

$$\lim_{n\to\infty }\left(1+\frac1n\right)^n\le \sum_{k=0}^\infty \frac1{k!}\tag2$$

We also have for $2\le m\le n$

$$\begin{align} \left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\ &\ge\sum_{k=0}^m \frac1{k!}\prod_{j=0}^{k-1} \left(1-\frac{j}{n}\right)\tag3 \end{align}$$

Letting $n\to\infty$ in $(3)$ reveals

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\ge \sum_{k=0}^m \frac1{k!}\tag4$$

whereupon letting $m\to \infty$ in $(4)$ yields

$$\lim_{n\to \infty}\left(1+\frac1n\right)^n\ge \sum_{k=0}^\infty \frac1{k!}\tag5$$

Putting together $(2)$ and $(5)$ provides proof of equality. And we are done!

Mark Viola
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