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.


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 $p>1$.

  • $\begingroup$ Is this correct? $\endgroup$ – mirgee Apr 5 '14 at 8:28
  • $\begingroup$ I'm pretty sure it is. $\endgroup$ – Marko Karbevski May 19 '14 at 18:42

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 $e^{n(\frac{1}{n}-\frac{1}{2n^2}+o(1/n^3))})$.

This clearly shows that the series converge for p>1 and diverge otherwise.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.