# Series convergence of $\sum_{n=1}^{\infty}\frac{1}{n^3}$

I have the infinte series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ which I believe converges.

As the ratio test proved inconclusive, I am trying to use the comparison test in order to prove it's convergence, but I am unsure what series to compare it to. Initially, I believed I could compare it to $$\frac{1}{n^2}$$ but I am also unsure how to prove that this series converges.

Is there a general method for choosing the series with which you compare your series to?

• As a general rule, series of this form can be bounded by the relevant integral.
– lulu
Apr 10 '19 at 13:10

Yes, you can use the fact that, for each natural number $$n$$, $$\dfrac1{n^3}\leqslant\dfrac1{n^2}$$. And then you can use the fact that $$\dfrac1{n^2}\leqslant\dfrac1{n(n-1)}$$ if $$n\neq1$$. Now, you have$$\sum_{n=2}^\infty\frac1{n(n-1)}=\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1n\right),$$and this series converges (it is a telescoping series).

But it is more natural to use the integral test to study the convergence of $$\displaystyle\sum_{n=1}^\infty\frac1{n^3}$$.

You can use that $$\frac{1}{n^3}\le \frac{1}{n^2}$$ and $$\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$$

You can also use integral test i.e testing whether $$\int_{1}^{\infty}{\frac{dx}{x^3}}$$ converges which seem to be easy to solve

$$\sum_{n=1}^\infty \frac1{n^s} \lt 1+ \int_1^\infty \frac1{x^s} dx=\frac{s}{s-1}$$ For all $$s\gt 1$$.

The answer is Apery's constant, or the limit of the Zeta function at $$s=3$$

https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant

There is no known way to evaluate this, but it does converge, as it is smaller than the series for $$s=2$$, whose proof is given in this beautiful video by 3Blue1Brown