This question is from Stein's Complex Analysis, and this is the first part of the original question. The original question was posted here but it was only about the remaining part, and I do not really have problems with the remaining part. So, here is the question I wanted to ask:
Let $F(z)=\sum_{n=1}^{\infty}d(n)z^{n}$ where $d(n)$ denotes the number of divisors of $n$. Observe that the radius of convergence of this series is $1$. Verify the identity:
$\sum_{n=1}^{\infty}d(n)z^{n}=\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$
The remaining part of the original question is to use this identity to show something else, but I don't know how to verify this identity..
I have tried some example for some n's, and I believe this identity, but I don't really have a clue about how to verify this identity. Here is what I tried so far:
$\frac{1}{1-z^{n}}=1+z^{n}+z^{2n}+z^{3n}+z^{4n}+\cdots$, and then $\frac{z^{n}}{1-z^{n}}=z^{n}+z^{2n}+z^{3n}+z^{4n}+z^{5n}+\cdots$.
So, we have $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}z^{kn}$.
Then I don't know how to do next.
Some hints or detailed explanations are really appreciated!!
You are nearly there. Consider how many ways $z^m$ appears is the double sum $$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}z^{kn}.$$ It appears once for each way of writing $m=kn$, that is once for each $k$ that's a positive integer factor of $m$. Thus the coefficient of $z^m$ is $d(m)$: the double sum is $$\sum_{m=1}^\infty d(m)z^m.$$