Is $\sum^n_{k=0}\frac{x^k}{k^2}$ Cauchy in $n$?

Question

I am extremely out of practice in figuring out convergence of sequences/series - I want to figure out if $$\sum^n_{k=0}\frac{x^k}{k^2}$$ is a Cauchy sequence in $n$ when $x\in [0,1]$, that is, if for each $\epsilon > 0$, there is an integer $N$ such that for all $n,m \ge N$, $$\sum_{k=0}^n \frac{x^k}{k^2} - \sum_{k=0}^m\frac{x^k}{k^2}.$$ So far, all I have is that if $m >n$, we have $$\sum_{k=n+1}^m \frac{x^k}{k^2},$$ and I've also thought of trying to look at when $x = 1$, which gives us $$\sum^m_{k=n+1}\frac{1}{k^2},$$ and maybe if that sequence converges, then the sequence will converge for all $x\in [0,1]$. Problem being that even though I think that $\sum^\infty_{k=1}\frac{1}{k^2}$ converges, I don't know how to handle the fact that the starting index is $0$.

Am I on the right track at all? Am I totally wrong? Thanks.

Answer 1

I think the starting index 0 is a typo.

We have: $\sum_{k=1}^{\infty} \frac{x^k}{k^2} \leq \sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}< \infty$. Your infinite sum is convergent, thus is Cauchy.

Answer 2

Recall that

$$\text{Li}_2(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^2}$$

for $|x|\le 1$ is the series representation of the dilogarithm function. The series converges on the interval of interest absolutely and uniformly.

Inasmuch as a sequence converges if and only if it is a Cauchy sequence, then convergence alone of the series implies that its partial sum constitutes a Cauchy sequence.