I was experimenting with a technique I developed for double/multiple summation problems, and thought of this problem:
Find $$S(p)=\sum_{n=1}^{\infty} \frac{\ln(n) \tau(n)}{n^p}$$ where $\tau(n)=\sum_{d|n} 1$ is the number of positive divisors of $n$.
Here's how I solved it:
Consider the sum $$S_2(p)=\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{\ln(ij)}{(ij)^p}$$ We will show that $S_2(p)=S(p)$. First, consider an arbitrary term $\frac{\ln(n)}{n^p}$. This exact term will appear(when double summation is expanded out) exactly $\tau(n)$ times, because $n$ can be written as $ij$ $\tau(n)$ different ways for $i,j \geq 1$. Therefore, $$S_2(p)=\sum_{n=1}^{\infty} \frac{\ln(n) \tau(n)}{n^p} = S(p)$$ as desired. Now, we work on evaluating $S_2(p)$ which is easier to work with.
To do this, we rewrite as follows: $$S_2(p)=\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \frac{\ln(i)+\ln(j)}{(ij)^p}=$$ $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} (\frac{\ln(i)}{(ij)^p}+\frac{\ln(j)}{(ij)^p})=$$ $$\sum_{i=1}^{\infty} (\sum_{j=1}^{\infty} \frac{\ln(i)}{(ij)^p}+ \sum_{j=1}^{\infty} \frac{\ln(j)}{(ij)^p})=$$ $$\sum_{i=1}^{\infty} (\frac{\zeta(p) \ln(i)}{i^p} + \frac{-\zeta'(p)}{i^p})=$$ $$- \zeta(p) \zeta'(p) - \zeta'(p) \zeta(p)=$$ $$\boxed{-2 \zeta(p) \zeta'(p)}=$$
and we're done.
I have a few questions regarding this problem. The first, obviously, is if my answer is correct. The second is if there is any simpler way to evaluate the sum. My method involves writing it as a double summation, but I was wondering if there are any straightforward single-sum ways to deal with this.
Your answer is correct.
You can use the following known fact:
$$\left(\sum_{n=1}^\infty \frac{f(n)}{n^s}\right)\left(\sum_{n=1}^\infty \frac{g(n)}{n^s}\right) = \sum_{n=1}^\infty\frac{(f*g)n}{n^s}$$ where $f*g$ is the Dirichlet convolution of $f$ and $g$ and observe that $\log \cdot \tau = 2\ (\log * 1)$.
But this is almost the same proof you gave, hiding the details in one well known theorem.