Another Divergence.

Algebra Level 4

For each positive integer $k$ , let $A(k)$ be the number of odd divisors of $k$ in the interval $\left[1, \sqrt{2k} \right)$ .

Evaluate:

$\large \sum _{ k=1 }^{ \infty }{ { \left( -1 \right) }^{ k-1 }\frac { A(k) }{ k } } .$

\frac { { \pi }^{ 2 } }{ 16 }

\frac { { \pi }^{ 2 } }{ 16\sqrt{2} }

Series is diverging.

\frac { 3{ \pi }^{ 2 } }{ 16 }

\frac { { \pi }^{ 2 } }{ 9 }

\frac { { \pi }^{ 2 } }{ 12\sqrt{2} }

\frac { { \pi }^{ 2 } }{ 12 }

1 solution

Mark Hennings
Nov 17, 2019

@Mark Hennings , i think you have solved all Putnam Problems!

Priyanshu Mishra - 1 year, 6 months ago