Convergence of $\sum_{j=1}^{\infty} \frac{1}{j^3+x^2}$

Question

I'm having some confusion with inferring the convergence of

$$\sum_{j=1}^{\infty} \frac{1}{j^3+x^2}$$

This looks like a good candidate for Weierstrass M-test.

However, if one does

$$\bigg | \frac{1}{j^3+x^2} \bigg | < \frac{1}{j^3} =: M_j$$

then one is lead to prove the second property required by Weierstrass M-test, which is

$$\sum_{j=1}^{\infty} M_j < \infty$$

But $$\sum_{n=1}^{\infty} \frac{1}{n^3}$$

is well-known to be the "non-trivial" Apéry's constant.

So is the above sum's value known "well enough" so that I can show/prove that

$$\sum_{j=1}^{\infty} M_j < \infty$$

or should I do something else with my series?

Answer 1

Of course it's correct what you have calculated.

To avoid confusions with Apery's constant:

$$\sum_{j=1}^{\infty} \frac{1}{j^3+x^2}\leq \sum_{j=1}^{\infty} \frac{1}{j^3}< \sum_{j=1}^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}$$

and $\sum_{j=1}^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}$ is well-known.

Answer 2

You don't need to know the precise value of the limit to know convergence. Since $1/j^3$ is decreasing, we can compare with an integral $$ \sum_{i=1}^\infty \frac1{j^3} = 1 + \sum_{i=2}^\infty \frac1{j^3} \le 1 + \int_1^\infty \frac1{t^3}\,dt = \frac32. $$ Alternatively, we can compare with the telescoping series $$ \sum_{i=1}^\infty \frac1{j^3} < 1 + \sum_{i=2}^\infty \frac1{j^3-j} = 1 + \frac12 \sum_{i=2}^\infty \biggl[ \frac{1}{j(j-1)} - \frac{1}{(j+1)j} \biggr] = \frac54. $$