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.