A question about series convergence

Question

Prove that $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$ diverges.

This is my proof :

I'm going to use Cauchy's creterion. Let $n>m$.

We have $lim_{n \longrightarrow \infty} (1+\frac{1}{n})^n=e$ thus $\exists n_1 \in \mathbb{N}$ such that $(1+\frac{1}{n})^n >e-1/2, \forall n \geqslant n_1$.

If the seiries connverges,then $\exists n_0 \in \mathbb{N} $ such that $n_0>2$
and $|(e-(1+\frac{1}{m})^m)+.....+(e-(1+\frac{1}{n})^n)|< 1$, $\forall n>m \geqslant n_0$

We see that $|(e-(1+\frac{1}{m})^m)+.....+(e-(1+\frac{1}{n})^n)|>(n+1-m)(e-(1+\frac{1}{n})^n)$

If we take $m=max\{n_0,n_1\}$ and $n=10(max\{n_0,n_1\})-1$ we deduse that

$9\max\{n_0,n_1\}/2 <1 $ deriving a contradiction

Is this proof valid or do i have to proceed with another aproach to this problem?

Answer 1

Note that, by using the Taylor's expansion $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$, we have that $$\left(1+\frac{1}{n}\right)^n=\exp\left(n\ln\left(1+\frac{1}{n}\right)\right)=\exp\left(n\left(\frac{1}{n}-\frac{1}{2n^2}+o(1/n^2)\right)\right)\\=e\cdot \exp\left(-\frac{1}{2n}+o(1/n)\right).$$ Hence, by the Taylor's expansion $e^x=1+x+o(x)$, $$e-\left(1+\frac{1}{n}\right)^n\sim e-e\cdot \exp\left(-\frac{1}{2n}\right)\sim \frac{e}{2n}.$$ which implies that the series is divergent.

Answer 2

I just felt like giving another proof of the results above. I claim

$$\lim\limits_{n\to \infty} n\left[e-\left(1+\frac{1}{n}\right)^n\right]=\frac{e}{2}$$

the series then diverges by a limit comparison with $\sum \frac{1}{n}$.

To show the above limit (it's well known isnt it ?) let $n=\frac{1}{x}$ and apply l'hopitals rule twice to

$$\frac{e-(1+x)^{\frac{1}{x}}}{x}.$$