# 2016 Putnam B6 difficult summation problem

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?

• A few analytical solutions here for problem B6.
– Syed
Oct 11, 2022 at 9:20

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: $$0<\Re(s)<1$$

• Mariusz Iwaniuk: Very nice... I'm guessing from a real mathematician who knows Mathematica rather than from a computer scientist. ($\checkmark$) Oct 11, 2022 at 12:23