# $\sum_{n=1}^{4000000} \frac{1}{n^3}$ very quick.

Some days ago I have tried to find the sum of the first milion terms of the infinite sum $$\zeta(3) = \sum_{n=1}^\infty\frac{1}{n^3}$$ (Apéry's constant) on Wolfram Programming Lab (Open Cloud), an online CAS. Therefore I wanted calculate: $$\sum_{n=1}^{1000000}\frac{1}{n^3}$$.

Since the calculation power available for free users is limited (and the ParallelSum function is not available online) I started evaluting s=Sum[1/(n^3),{n,1,100000}](the first one hundred thousand terms). Then s=Sum[1/(n^3),{n,100001,200000}] (the second one hundred thousand terms) and s=Sum[1/(n^3),{n,200001,300000}] (the third one hundred thousand terms). Now the calculation Sum[1/(n^3),{n,300001,400000}] was aborted by the system because it exceeded the time limit. I think that this calculations are very hard to do and all the previous calculations have required some time for the output on this very fast system. Now I have input s=Sum[1/(n^3),{n,300001,350000}] and the output was come out. And now for me was happen a strange thing. I make a typing error and I have input s=Sum[1/(n^3),{n,350001,4000000}] (4 milions!), but the system very quickly output the following answer: (1/2)(-PolyGamma[2,350001]+PolyGamma[2,4000001]

Is it mathematically correct? I have make the sum N[s+s+s+s+s] and compared with the know value of the Apéry's constant and it seems correct.

• Therefore is it correct? – vi pa Sep 24 '18 at 14:47

It is mathematically correct. See Eqn (11) in MathWorld's Polygamma article. Accordingly, the result is \begin{align} \tfrac{1}{2}(\psi_2 (4000001)-\psi_2 (3500001))&=\tfrac{1}{2}(-1)^3 2!\left( \zeta(3)-H^3_{4000000} - \zeta(3) + H^3_{3500000} \right)\\ &=-\left(\sum_{n=1}^{4000000}\frac{1}{n^3}-\sum_{n=1}^{3500000}\frac{1}{n^3}\right) \\ &=\sum_{n=3500001}^{4000000}\frac{1}{n^3} \end{align} Mathematica decided that it was more useful to give you the Polygamma version of the sum rather than carry out the entire computation.