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 \(a+b+c\).


  • 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 \(d^{3}*1 = (d*1)^{2}\) .


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