After that I was playing with Wolfram Alpha online calculator around the Dirichlet series of the sum of divisors function $$\sigma(n)=\sum_{d\mid n}d,$$ (my idea was to explore series involving the Apéry's constant, but seems failed) I wondered what about of the series $$\sum_{n=1}^\infty\frac{\sigma(n^2)}{n^6},\tag{1}$$ and if is it is possible other generalizations with a good mathematical meaning.

Using the CAS I've found a remarkable coincidence with a rational multiple of a particular value of the Riemann zeta function, but I don't know a proof (I know the method for multiplicative functions using Euler's products).

Question. I don't know if the closed-form that I've evoked for $(1)$ was in the litearute. Were in the literature the series $(1)$ or similar generazalitions? Then please refer it, and I try to find and read those propositions. Many thanks.

Notice that I don't know neither if it was in the literature nor how if the formaula that I've evokes it is easy to prove. Thus I a requiring this reference request.


More generally, you have $$\sum\limits_{n = 1}^\infty {\frac{{{\sigma _a}({n^2})}}{{{n^s}}}} = \frac{{\zeta (s)\zeta (s - a)\zeta (s - 2a)}}{{\zeta (2s - 2a)}}$$

Indeed, $$\sum\limits_{n = 1}^\infty {\frac{{{\sigma _a}({n^2})}}{{{n^s}}}} = \prod\limits_p {\left( {1 + \frac{{{\sigma _a}({p^2})}}{{{p^s}}} + \frac{{{\sigma _a}({p^4})}}{{{p^{2s}}}} + ...} \right)} $$ and the result follows at once by calculating the inner term (it's just a geometric series): $$\sum\limits_{k = 0}^\infty {\frac{{{\sigma _a}({p^{2k}})}}{{{p^{ks}}}}} = \frac{{1 + {p^{ - (s - a)}}}}{{(1 - {p^{ - (s - 2a)}})(1 - {p^{ - s}})}}$$

  • $\begingroup$ Many thanks then I am going to do the exercise. $\endgroup$ – user243301 Feb 9 '18 at 16:12

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