# q-series related to d(n)=# of divisors of n

Let $\sigma_m (n)=\sum_{d|n}d^m$, and $d(n)=\sigma_0(n)$ as stated in the title.

My question is: Does the q-series $f(q)=\sum_{n\ge 1} d(n)q^n$ gives something like a modular form (i.e. with some kind of identity under the action of $SL_2$) when $q=e^{2\pi i z}$?

When $d(n)$ is replaced by $\sigma_{2m-1} (n)$ the q-series gives out Eisenstein series, which are exactly modular forms. One way to obtain such a relation is to use Hecke's converse theorem for the L-function $\zeta(s) \zeta(s-2m+1)$. So I feel natural to think about $\zeta(s)^2$ for properties of $\sum_{n\ge 1} d(n)q^n$.

I tried to modify the argument of Hecke's converse theorem but it does not really give me a neat relation.

Any help will be appreciated. Thanks in advance.

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The coefficient $d(n)$ is the limit of the $n$-th Hecke eigenvalue of the (nonholomorphic) Eisenstein series $E(z,s)$ as $s\to 1/2$. The limit of the Eisenstein series itself is zero, hence it is more natural to consider the $(\partial/\partial s)E(z,s)$ at $s=1/2$, which is precisely $$\sqrt{y}\log y+4\sqrt{y}\sum_{n=1}^\infty d(n)K_0(2\pi ny)\cos(2\pi nx).$$
In short, you have to leave the realm of holomorphic modular forms to find the object you are looking for: $d(n)$ is "morally" the $n$-th Hecke eigenvalue of the Eisenstein series with Laplace eigenvalue $1/4$ (which is zero eventually, so we pass to the derivative). This is another reason why Maass forms are so natural, something that not every arithmetician appreciates!