Solving a weird series

Number Theory Level 5

$\sum_{n=1}^\infty \dfrac{d(n) }{n^2}$

Let $d(n)$ denote the number of positive divisors of integer $n$ inclusive of 1 and itself.

If the series above is equal to $\dfrac {a \pi^b}c$ , where $a,b,c$ are positive integers with $a,c$ coprime, find the value of $a+b+c$ .

1 solution

Samarpit Swain
Jan 1, 2016

The sum is one of the Dirichlet series, given by: $\sum_{n=1}^\infty \dfrac{d_{x}(n) }{n^s}=\zeta{(s)}\zeta{(s-x)}$

In this case $x=0$ and $s=2$ , $\therefore \sum_{n=1}^\infty \dfrac{d(n) }{n^2}=\zeta(2)^2=\dfrac{\pi^4}{36}$

How did you find zeta(2)?

Aryan Gaikwad - 5 years, 3 months ago

It is a well known result.

Sal Gard - 5 years, 1 month ago

i don't know how did you found the x :(

aaaaaaa aaaaaa - 4 years, 9 months ago