Uniformly convergent series on $\vert x\vert \leq k$ where $k\in (0,1)$

Question

I want show that each of the series $\sum^{\infty}_{n=1}\frac{nx^n}{1-x^n}$ and $\sum^{\infty}_{n=1}\frac{x^n}{(1-x^n)^2}$ converges uniformly on $\vert x\vert \leq k$ where $k\in (0,1)$; and prove that the two series are equal. I tried to apply the Weierstrass M-test but it didn't work for me. Could somebody give me a hint ?

Answer 1

Assuming $|x|<1$, we have: $$f(x)=\sum_{n=1}^{+\infty}\frac{nx^n}{1-x^n}=\sum_{n=1}^{+\infty}n\left(x^n+x^{2n}+x^{3n}+\ldots\right) = \sum_{n=1}^{+\infty}\sigma_1(n)\,x^n$$ with $\sigma_1(n)=\sum_{d\mid n}d<n^2$. This gives that the radius of convergence is one and $f(z)$ is a holomorphic function on the open ball $|z|<1$. The situation is the same for: $$ g(x)=\sum_{n=1}^{+\infty}\frac{x^n}{(1-x^n)^2}=\sum_{n=1}^{+\infty}\left(x^n+2 x^{2n}+3x^{3n}+\ldots\right)= \sum_{n=1}^{+\infty}\sigma_1(n)\, x^n.$$ It is interesting to notice that over $|z|<1$ we have: $$ g\equiv f. $$

Answer 2

Hint. In both cases denominator $\rightarrow 1$ hence you can replace it for instance with $\frac12$ and show that new series are also converges. Finally you have to show that new series multiplied by some constant majorize yours.

Since $1 - x^n \to 1$ then there exists some $N$ that $\forall n>N:1 - x^n > \frac12$ (finite number of summands doesn't matter). Hence $$\forall n>N:\frac{nx^n}{1-x^n} <2nx^n$$

Similarly with second series. Finally, you can majorize $2nx^n < 2nk^n$ and this series are converges since $k \in (0,1)$. I think it is Weierstrass M-test, isn't it?