I want to prove that $\sum\limits^{\infty}_{n=1} \frac {2^n} {n^n}$ is convergent. Therefore I want to use that it will be convergent if $\lim\limits_{n \to \infty}|\frac {2^{n+1}} {n^n}| < 1$ is true.

But I am not sure if all of my steps are legal to do in order to do it. Here is what I came up with:

$$\lim_{n \to \infty}|\frac {2^{n+1}} {n^n}| < 1$$ $$\lim_{n \to \infty}|\frac {\sqrt[n]{2^{n}} \sqrt[n]{2}} {\sqrt[n]{n^n}}| < \sqrt[n]1$$ $$\lim_{n \to \infty} \frac {2 \sqrt[n]2} {n} < 1$$ $$0<1$$

$\Rightarrow$ It is true.

  • $\begingroup$ It look fine, but you could easily use root test en.wikipedia.org/wiki/Root_test $\endgroup$ – arberavdullahu Jan 9 '17 at 9:18
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    $\begingroup$ What convergence test are you using? $\endgroup$ – Fabian Jan 9 '17 at 9:20
  • $\begingroup$ Your proof is very far from being "fine". First major trouble: "I want to use that ... $\sum\limits^{\infty}_{n=1} \frac {2^n} {n^n}$ ...(is) convergent if $\lim\limits_{n \to \infty}|\frac {2^{n+1}} {n^n}| < 1$" Sorry but how does the condition that $\lim x_n<1$ with $(x_n)$ positive, supposed to imply that the series $\sum x_n$ converges? $\endgroup$ – Did Jan 9 '17 at 10:21
  • $\begingroup$ ((Four users cannot be bothered to read the title, apparently.)) $\endgroup$ – Did Jan 9 '17 at 10:23
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    $\begingroup$ What you "want to use" seems to be wrong, as that'd give us that as $\;\lim\limits_{n\to\infty}\frac1n=0\;$ then the harmonic series is convergent. It is true the limit you want to calculate is less than one...in fact it is zero, but that is far from being a sufficient condition for a series to converge. $\endgroup$ – DonAntonio Jan 9 '17 at 10:26

As commented before $$\lim_{n\to +\infty}\sqrt[n]{\frac{2^n}{n^n}}=\lim_{n\to +\infty}\frac{2}{n}=0<1\;\underbrace{\Rightarrow}_{\text{Root test}} \;\sum_{n=1}^{+\infty}\frac {2^n} {n^n}\text{ is convergent.}$$


Of course, Alone's proof is striking and adapted. Below a proof which invites you to feel the growth rates.

For $n\geq 3$ $$ \frac {2^n} {n^n}\leq (\frac {2} {3})^n $$ and you can use the "direct comparison test" for series here.

  • $\begingroup$ More simple and intelligent. +1 $\endgroup$ – Fernando Revilla Jan 9 '17 at 17:01
  • $\begingroup$ @Alone thks a lot $\endgroup$ – Duchamp Gérard H. E. Jan 10 '17 at 1:36

The series is increasing and bounded:


As the terms decrease very quickly (faster than an exponential), few of them are necessary for a good estimate. For instance, with the first $9$ ones, the remainder is bounded by



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