One more of those

S = n = 1 σ 2 ( n ) n 6 S=\sum_{n=1}^{\infty}\frac{\sigma_2(n)}{n^6}

Let σ 2 ( n ) \sigma_2(n) denote the sum of the squares of all the positive integer divisors of n n . For example, σ 2 ( 6 ) = 1 2 + 2 2 + 3 2 + 6 2 = 50 \sigma_2(6)=1^2+2^2+3^2+6^2=50 .

Enter π 10 S \frac{\pi^{10}}{S} as your answer.


The answer is 85050.

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1 solution

Aareyan Manzoor
Jan 5, 2016

read about dirichlet convolution (still under construction, if you want to help DM me at slack)

the defination of dirchlet series D G ( F ; s ) = n = 1 F ( n ) n s DG(F;s)=\sum_{n=1}^\infty \dfrac{F(n)}{n^s} where F is an arithmetic function. we are to know D G ( f g ; s ) = D ( f ; s ) D ( g ; s ) DG(f*g;s)=D(f;s)D(g;s) where f g f*g denotes the dirichlet convolution of f and g. define function f ( n ) = n 2 f(n)=n^2 and u ( n ) = 1 u(n)=1 . so we see that f u = d n ( f ( d ) u ( n d ) ) f*u=\sum_{d|n}\left(f(d)u\left(\dfrac{n}{d}\right)\right) but u of anything is one so: f u = d n ( d 2 ) = σ 2 ( n ) f*u=\sum_{d|n}\left(d^2\right)=\sigma_2(n) so D G ( σ 2 ; 6 ) = D G ( f ; 6 ) D G ( u ; 6 ) = ( n = 1 n 2 n 6 ) ( n = 1 1 n 6 ) = ζ ( 4 ) ζ ( 6 ) DG(\sigma_2;6)=DG(f;6)DG(u;6)=\left(\sum_{n=1}^\infty \dfrac{n^2}{n^6}\right)\left(\sum_{n=1}^\infty \dfrac{1}{n^6}\right)=\zeta(4)\zeta(6) we know we can calculate zeta of even integers using Bernoulli numbers.so, this equals after simplification π 10 85050 \dfrac{\pi^{10}}{\boxed{85050}}

Wonderful! That's exactly what I had in mind! (+1)

One small complaint, though: You butchered the name of my beloved country man (Jakob) Bernoulli, a member of the greatest mathematical family in history. ;)

Otto Bretscher - 5 years, 5 months ago

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Hm... I tried making a generating function but got stuck.

Alan Yan - 5 years, 5 months ago

sorry for that....... i have corrected it.....

Aareyan Manzoor - 5 years, 5 months ago

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