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Looking to prove that the following series converge or diverge. This is my first time attempting to use the root test and I am not sure if I proceeded correctly or not.

$$\sum \frac{1}{n^n}$$

$$\lim_{n\to\infty} |\frac{1}{n^n}|^{\frac {1}{n}} = \lim_{n\to\infty} (\frac {1}{n^n})^\frac {1}{n} = \lim_{n\to\infty}\frac {1^{\frac{1}{n}}}{n^\frac{n}{n}} = \lim_{n\to\infty}\frac {1}{n} =0 <1.$$ Therefore this series must converge

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    Do you **have** to use the root test? If not, comparison is extremely easy. That said, your work looks fine for this one--well done! – 2017-02-07
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    Actually you have to calculate the $\limsup_n$ but since the limit exists it is equivalent. You lost one $\lim$. The proof is okay. – 2017-02-07
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    Looks good. If I may note, one could also do the following comparison test:$$\frac1{n^n}<\frac1{n^2}$$for all $n>2$. – 2017-02-07
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    I do not have to, but I figured this would be a good way to go about solving the question – 2017-02-07
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    could you submit a answer using the comparion test? I would like to see how it works – 2017-02-07

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Yes. Your proof is valid

Your answer definitely works here. There is one slight caveat in that you should be calculating the Limit Superior, but it's fine to use a plain limit here because the plain limit exists.

If you want to use the Direct Comparison Test note that $$n^n > n^2 \implies \frac{1}{n^n} < \frac{1}{n^2}$$
We now know that $$\sum_{n=1}^\infty \frac{1}{n^n} < \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
If you want something a bit easier, you might try $n^n > 2^n$ as @Omnomnomnom notes $$\sum_{n=1}^\infty \frac{1}{n^n} < \sum_{k=1}^\infty \frac{1}{2^n} = 1$$ Where we used the Geometric Series Formula

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    do I need to calculate the limit superior? – 2017-02-07
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    @RT123 If the limit doesn't exist, yes, you need to use $\limsup$ instead. – 2017-02-07
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    @RT123 if you are in a calculus class and have never heard the term "limit superior" in class, then no. – 2017-02-07
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    $n^n > 2^n$ strikes me as the more obvious comparison here – 2017-02-07
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    so the limit does exist in this question so the root test is fine? – 2017-02-07
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    @RT123 the limit exists, so using *the version of the root test with a limit instead of a limit superior* is fine. The $\limsup$ is only something that one might worry about in the case where $\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}$ fails to exist. – 2017-02-07
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    @Omnomnomnom Fair enough :) I think it's easier to see that $n^n >n^2$ because the LHS is just $n^2 n^{n-2}$ and it should be clear that, for sufficiently large $n$, $n^{n-2}$ is positive – 2017-02-07
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    @Brevan and I think it's easy to see that $n^n > 2^n$ since the "$n$th root function" is increasing. To each his own, I suppose. – 2017-02-07
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    @Omnomnomnom what I will agree with you on is that calculating the sum of $2^{-n}$ is easier for a beginner, so I made a note on that. – 2017-02-07
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Since $\frac{1}{n^n} \leq \frac{1}{n^2}$ for all $n\in \mathbf N$ and $\sum_n \frac{1}{n^2}$ is convergent the series $\sum_n \frac{1}{n^n}$ is convergent aswell.

Beside all these different methods you can use the ratio test aswell if you like. Set $a_n =\frac{1}{n^n}$. Then we have $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{n^n}{(n+1)^{n+1}}=\lim_{n\to \infty} \frac{1}{n+1} \frac{n+1}{n}\left(1- \frac{1}{n+1} \right)^{n+1}=0,$$ where we used that $$ \lim_{n\to \infty} \frac{1}{n+1}=0, \qquad \lim_{n\to \infty} \frac{n+1}{n}=1 \qquad \text{and} \qquad \lim_{n\to \infty}\left(1- \frac{1}{n+1} \right)^{n+1}=\mathrm e^{-1}$$