Problem B6 on the 2016 Putnam exam is to calculate:
$$\sum\limits_{k=1}^\infty \left( \frac{(-1)^{k-1}}{k} \sum\limits_{n=0}^\infty \frac{1}{k 2^n + 1} \right)$$
The direct approach
Sum[(-1)^{k-1}/k Sum[1/(k 2^n + 1),{n,0,\[Infinity]}], {k,1, \[Infinity]}]
does not resolve to the analytic solution. However, the numerical value, computed by N[%] gives the correct value: $1$.
How can we compute the analytic solution to this summation?
One way using MellinTransform and InverseMellinTransform:
func = (-1)^(k - 1)/k*1/(A*k 2^n + 1)(* where A = 1*)
InverseMellinTransform[Sum[Sum[MellinTransform[func, A, s] //
PowerExpand, {k, 1, Infinity}], {n, 0, Infinity}], s, A] /. A -> 1
([Pi] InverseMellinTransform[Csc[[Pi] s] Zeta[1 + s], s, 1])(Can't compute ! Weakness !)
Using:
$$\zeta (1+s)=\sum _{k=1}^{\infty } \frac{1}{k^{1+s}}$$
Then:
InverseMellinTransform[\[Pi] Csc[\[Pi] s] 1/k^(1 + s), s, A] /. A -> 1 // FullSimplify
(*1/(k + k^2)*)
Sum[%, {k, 1, Infinity}]
(* 1 *)
Addition for Table of InverseMellinTransform: $$\mathcal{M}_s^{-1}[\pi \csc (\pi s) \zeta (1+s)](A)=\gamma +\psi ^{(0)}\left(1+\frac{1}{A}\right)=H_{\frac{1}{A}}$$ for: $0<\Re(s)<1$
$$\mathcal{M}_s^{-1}[\pi \csc (\pi s) \zeta (2-s)](A)=\frac{\gamma }{A}+\frac{\psi ^{(0)}(1+A)}{A}=\frac{H_A}{A}$$
for: $0<\Re(s)<1$
