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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.

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    It look fine, but you could easily use root test https://en.wikipedia.org/wiki/Root_test – 2017-01-09
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    What convergence test are you using? – 2017-01-09
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    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? – 2017-01-09
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    ((Four users cannot be bothered to read the title, apparently.)) – 2017-01-09
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    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. – 2017-01-09
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    If $ 0< \lim \frac{2^n}{n^n}=a<1$, the summation would increase forever. – 2017-01-09

3 Answers 3

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

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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.

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    More simple and intelligent. +1 – 2017-01-09
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    @Alone thks a lot – 2017-01-10
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The series is increasing and bounded:

$$\sum_{n=1}^{\infty}\left(\frac2n\right)^n<2+1+\sum_{n=3}^{\infty}\left(\frac23\right)^n=\frac{35}9.$$

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

$$\sum_{n=10}^{\infty}\left(\frac2{10}\right)^n=\frac1{5^9\cdot4}=0.000000128.$$