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Tried using $b_n = \frac1{n^n + 1}$ with limit test which indicated that both either converge or diverge but getting stuck on how to show that one actually does converge.

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Use that $\sum\limits_{k=1}^{\infty} a_k $ converges if and only if $\sum\limits_{k=j}^{\infty} a_k $ converges for a $j \in \mathbb N$

Edit: Here is the full solution:

$\sum\limits_{n=1}^{\infty} \frac{1}{n^n}<\infty \iff \sum\limits_{n=2}^{\infty}

\frac{1}{n^n}<\infty$

and $ \sum\limits_{n=2}^{\infty} \frac{1}{n^n}=\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}+…. \leq \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+….=\sum\limits_{n=2}^{\infty} \frac{1}{n^2}< \infty$

You can conclude if the series converges or not,using the ratio test.

It is like that:

$a_n=\frac{1}{n^n}$

$$\frac{a_{n+1}}{a_n}=\frac{\frac{1}{(n+1)^{n+1}}}{\frac{1}{n^n}}=\frac{n^n}{(n+1)^{n+1}} \to 0<1$$

So,the series $\sum_{n=1}^{\infty} \frac{1}{n^n}$ converges.

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