Dirichlet It Everytime

Number Theory Level 5

$\displaystyle \sum_{n=1}^\infty \dfrac{\sigma_{2016} (n)}{n^{M}}$

Let $\sigma_A(n)$ denote the sum of $A^\text{th}$ powers of all the positive integer divisors of $n$ . For example, $\sigma_5(6) = 1^5 + 2^5 + 3^5 + 6^5 = 8052$ . Find the infimum of $M$ such that the series above converges.

Inspiration .

The answer is 2017.

1 solution

Samrat Mukhopadhyay
Sep 17, 2016

Note that we can write the sum as $\sum_{n\ge 1}\sum_{d|n}\frac{d^k}{n^{M}}$ where $k=2016$ . We can re write the sum as $\sum_{d\ge 1}\sum_{l\ge 1}\frac{d^k}{(ld)^M}=\sum_{d\ge 1}\frac{1}{d^{M-k}}\sum_{l\ge 1}\frac{1}{l^M}=\zeta(M-k)\zeta(M)$ Note that the sum converges as long as $M>k+1$ . So, we get $\inf M=k+1=\boxed{2017}$ .

hey, we got a forum where most of the regular users on site discuss about math and science, would u be interested to join?

Pi Han Goh - 4 years, 9 months ago

Yeah, of course.

Samrat Mukhopadhyay - 4 years, 9 months ago

Go here

Pi Han Goh - 4 years, 9 months ago

@Pi Han Goh – Thanks a lot!

Samrat Mukhopadhyay - 4 years, 9 months ago

Dirichlet It Everytime

The answer is 2017.

1 solution

0 pending reports