Divisor Generating Function?

Number Theory Level 5

It is known that $\large \displaystyle \sum _{n=1}^{\infty }\dfrac{\sigma_3(n)}{n^2}e^{-2\pi n}=\dfrac{G}{a} - \dfrac{b\pi^c}{d}$

where $G$ is denotes the Catalan's constant

$\displaystyle G=\sum _{n=0}^{ \infty }\frac{\left(-1\right)^n}{\left(2n+1\right)^2},$

$\sigma_3(n)$ is defined to be the divisor function

$\displaystyle σ_3(n)=\sum_{d\mid n}d^3,$

And $a,b,c$ and $d$ are positive integers , with $b,d$ coprime. Find $a+b+c+d$ .

Inspiration .

The answer is 739.

1 solution

Mark Hennings
Jul 31, 2016

Giving the details of your comment to my problem... Since $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{\sigma_3(n)}{n^2}e^{-2\pi n} & = & \displaystyle \sum_{n=1}^\infty \frac{e^{-2\pi n}}{n^2} \sum_{d|n} d^3 \; = \; \sum_{n=1}^\infty ne^{-2\pi n} \sum_{d|n} \frac{1}{d^3} \\ & = & \displaystyle \sum_{d=1}^\infty \frac{1}{d^3} \left(\sum_{n=1}^\infty (nd) e^{-2\pi nd}\right) \; = \; \sum_{d=1}^\infty \frac{1}{d^2} \frac{e^{-2\pi d}}{(1 - e^{-2\pi d})^2} \\ & = & \displaystyle \tfrac14\sum_{d=1}^\infty \frac{\mathrm{cosech}^2 \pi d}{d^2} \; = \; \tfrac14\sum_{d=1}^\infty \frac{\coth^2\pi d}{d^2} - \tfrac{1}{24}\pi^2 \\ & = & \displaystyle \tfrac14\left(\tfrac23G + \tfrac{19}{180}\pi^2\right) - \tfrac{1}{24}\pi^2 \; = \; \tfrac16G - \tfrac{11}{720}\pi^2 \end{array}$ using the answer to my problem, the result is $6 + 11 + 2 + 720 = \boxed{739}$ .

Divisor Generating Function?

The answer is 739.

1 solution

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