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
On
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 $$ \lim_{n\to \infty} \frac{1}{n+1}=0, \qquad \lim_{n\to \infty} \frac{n+1}{n}=1 \qquad \text{and} \qquad \lim_{n\to \infty}\left(1- \frac{1}{n+1} \right)^{n+1}=\mathrm e^{-1}$$
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