Does $\sum_{n=1}^\infty \frac{1}{n^{(n^2)}}$ convergent?

Question

For $$\sum_{n=1}^\infty \frac{1}{n^{(n^2)}}$$, can we use p-sereis theorem? To be more specifically, as n>=1, n^2>=1, so we can conclude that this series is convergent.

Answer 1

Since $$ \limsup_{n\to\infty}\sqrt[n]{\frac{1}{n^{n^2}}}=\limsup_{n\to\infty}\frac{1}{n^n}=0,$$ The root test tells us that the series in question is convergent.

Answer 2

Yes you can use p-series

$$\sum_{n=1}^\infty \frac{1}{n^{(n^2)}} < 1 + \sum_{n=2}^\infty \frac{1}{n^{2}}$$

Then use p-series theorem.

Answer 3

You can use $p$-series along with a comparison. You know that $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the $p$-series test. Then since the summand $\frac{1}{n^{n^2}}$ is always positive and $n \ge 1 \implies \frac{1}{n^{n^2}} \le \frac{1}{n^2}$, you get convergence of your series.