# $$\ln(n)$$ can make a difference

Calculus Level 5

the sum $\sum_{n=1}^\infty \dfrac{\tau(n)}{n^2}=\dfrac{\pi^4}{36}$ This has a very simple closed form, and is easy to work out.

However, the sum: $\sum_{n=1}^\infty \dfrac{\ln(n)\tau(n)}{n^2}$ Has a very difficult closed form, of the form $-\dfrac{\pi^a}{b}(c\gamma+\ln(d\pi)+f\ln(A))$ Find $$a+b+c+d+f$$.

Notations:

All of $$a,b,c,d,f$$ are integers.

$$\tau(n)$$ means the number of positive divisors of n.

A is the Glaisher-Kinkelin Constant

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