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Let $e_n$ be $(1+\frac{1}{n})^{\frac{1}{n}}$.
Let $E_n$ be $\sum_{k=0}^{n} \frac{1}{k!}$.
Suppose the convergence of $(e_n)$ and $(E_n)$ are established and let $e =lim_{n\to\infty}(e_n)$.

By Binomial Theorem,
$$e_n = 1+1+\frac{1}{2!}(1-\frac{1}{n})+\ldots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\ldots(1-\frac{n-1}{n})$$
Let $\epsilon >0$.
Since $e =lim_{n\to\infty}(e_n)$, there exists $N \in \mathbb{N}$ such that for each $n\ge N$, $|e_n-e| < \epsilon /2$. I was trying to show that there exists $N_2 \in \mathbb{N}$ such that $|E_n-e_n| < \epsilon /2$, but I failed.

How to prove that the two sequences converge to the same limit?

user387597
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1 Answers1

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Using Binomial series on $$\left(1+\dfrac1n\right)^n$$

the $r(0\le r\le n)$th term $$T_{r+1}=\dfrac{n(n-1)\cdots(n-r+1)}{r!n^r}=\dfrac1{r!}\prod_{u=0}^{r-1}\left(1-\dfrac un\right)$$

$$\lim_{n\to\infty}T_{r+1}=?$$

lab bhattacharjee
  • 274,582
  • This lacks rigor inasmuch as you have not shown that $$\begin{align}\lim_{n\to \infty}\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}&=1+\lim_{n\to \infty}\sum_{k=1}^n \prod_{\ell=0}^{k-1}\left(1-\frac\ell n\right) \frac{1}{k!}\\&=1+\sum_{k=1}^\infty \lim_{n\to \infty}\prod_{\ell=0}^{k-1}\left(1-\frac\ell n\right) \frac{1}{k!}\end{align}$$ – Mark Viola Jun 14 '17 at 18:44
  • Thank you for the hints. I think I have found an answer here, link – user387597 Jun 14 '17 at 19:09