# Decide convergence of the series

Using Taylor expansion decide convergence of the series:

$$\sum_{n=1}^{\infty}(e-(1+{{1}\over{n}})^n)^p = \sum_{n=1}^{\infty}a_n$$

I expanded $a_n$ like this

$a_n = (e-(1+{{1}\over{n}})^n)^p = (\sum_{k=0}^{\infty}{{1}\over k!} - \sum_{k=0}^{n}\dbinom{n}{k}{1\over n^k})^p = {1\over n^p}(const.+O({1\over n}))^p$

and then used the ratio test. Is it correct? If so, is there a better approach?

Thank you.

• – Lucian Apr 4 '14 at 22:39

Realize that $(1+{{1}\over{n}})^n = \sum_{k=0}^{n}\dbinom{n}{k}{1\over n^k} = {1\over k!}(1 - {1\over n})(1-{2\over n})...(1-{k-1\over n})< \sum_{k=0}^{n}({1\over k!}-{1\over nk!})$
We can expand $e-(1+{{1}\over{n}})^n$ as $\sum_{n=1}^{\infty}{1\over k!}(1 - (1 - {1\over n})(1-{2\over n})...(1-{k-1\over n}))> \sum_{k=0}^{n}{1\over nk!}>{2\over n}$. Therefore, $\sum_{n=1}^{\infty}(e-(1+{{1}\over{n}})^n)^p$ diverges when $\sum_{n=1}^{\infty}({2\over n})^p$ diverges, which is for $p \leq 1$. On the other hand, $e-(1+{{1}\over{n}})^n < (1+{{1}\over{n}})^n-(1+{{1}\over{n}})^{n+1}= {(1+{{1}\over{n}})^n \over n} \leq {e\over n}$. So $\sum_{n=1}^{\infty}(e-(1+{{1}\over{n}})^n)^p$ converges if $\sum_{n=1}^{\infty}({e\over n})^p$ converges, which is for $p>1$.
Rewrite the general term as $(e-e^{n\ln(1+1/n)})^p \sim (e-e^{n(\frac{1}{n}-\frac{1}{2n^2}+o(1/n^3))})^p =(e-e^{1-\frac{1}{2n}+o(1/n^2)})^p \sim e^p(1-(1+\frac{1}{2n }+o(1/n^2)+ o(X))^p\sim e^p\frac{1}{2^pn^p}$
$o(X)$ is the Taylor expansion rest of $e^{n(\frac{1}{n}-\frac{1}{2n^2}+o(1/n^3))})$.