Testing for convergence of expression with variable denominator raised to some constant 'a'
Solution 1:
Your series represents what we call genearlized harmonic serie that is defined as follows:
$$\sum_{n=1}^{+\infty}\frac{1}{n^{\alpha}}$$
Using Mengoli's serie, we can show that: $\alpha\in(1,+\infty)$ converges.
$\alpha\in (-\infty, 1]$ diverges.
Now, you can use the comparison test, to show that you series converges.
Comparison test. Let $a_n$ and $b_n$ two real sequences definitely non-negative and such that $1\leq a_n\leq b_n$. The following expression are true:
If $\sum a_n$ diverges, than also $\sum b_n$ diverges.
If $\sum b_n$ converges, then also $\sum a_n$ converges.