# Can you find the other way?

$\large\sum_{n=1}^\infty \dfrac{ \sigma_3 (n) }{n^6} = \dfrac{\zeta(a) \cdot \pi^b}c$

If the equation above holds true for positive integers $$a,b$$ and $$c$$, find $$a+b+c$$.

Notations:

• The divisor function for integer $$n$$, $$\sigma_k (n)$$ is defined as the sum of the $$k^\text{th}$$ powers of the positive integers divisors of $$n$$, $$\displaystyle \sigma_k(n) = \sum_{d| n} d^k$$.

• $$\zeta(\cdot)$$ denotes the Riemann zeta function.

###### I have posted one similar problem like this , but I think this problem has slightly different solutions.

Try to solve with $$d^{3}*1 = (d*1)^{2}$$ .

×