Where d(n) is the number of divisors of n, show that
$\sum_{n=1}^\infty d(n)z^n = \sum_{k=1}^\infty \frac{z^k}{1-z^k}$
where both sides converge for |z|<1
and show that
$\sum_{n=1}^\infty \frac{d(n)}{n^s} = (ζ(s))^2$
I've tried playing around with various forms of the divisor function and the Riemann Zeta function, but I can't seem to get anywhere with the first part of the problem. I would really appreciate some help getting started
On
The first one is pretty simple if we expand the geometric series: $$ \sum_{k=1}^\infty\frac{z^k}{1-z^k}=\sum_{k=1}^\infty\sum_{j=1}^\infty z^{jk}\tag{1} $$ Thus, $z^n$ appears in $(1)$ with a coefficient which is the number of ways we can write $n=jk$, which is $d(n)$.
The second one is a little trickier $$ \begin{align} \zeta(s)^2 &=\prod_{p\in\mathbb{P}}\frac1{(1-p^{-s})^2}\\ &=\prod_{p\in\mathbb{P}}\sum_{k=1}^\infty(k+1)p^{-ks}\tag{2} \end{align} $$ If $n=\prod\limits_{p\in\mathbb{P}}p^{e_p}$, then $d(n)=\prod\limits_{p\in\mathbb{P}}(e_p+1)$. Computing the coefficient of $n^{-s}$ in $(2)$ gives $d(n)$, therefore, $$ \zeta(s)^2=\sum_{n=1}^\infty\frac{d(n)}{n^s}\tag{3} $$
In $|z|<1$, $$\sum_{k\geq 1}\frac{z^k}{1-z^k}=\sum_{k\geq 1}\sum_{j\geq 1}z^{kj} = \sum_{n\geq 1} d(n)\, z^n.$$ For almost the same reason: $$\zeta(s)^2=\sum_{n,m\geq 1}\frac{1}{(nm)^s}=\sum_{k\geq 1}\frac{d(k)}{k^s}.$$