I would like to know how to calculate the sum of this convergent series

$$\sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}} $$

I cannot obtain this sum, mainly because it is not similar to any series of exponential, logarithmic or trigonometric functions, which can serve as a point of support.

Note: I would also like the sum not to be expressed in terms of complex numbers

Thanks in advance.

  • $\begingroup$ looks like you need to do more research, it easy to state hard problems that the solution is unknown, go look at zeta function for odd numbers. $\endgroup$
    – jimjim
    Commented Aug 16, 2020 at 0:17
  • $\begingroup$ @jimjim thanks, but in the course I'm developing they don't allow me to use the zeta function $\endgroup$ Commented Aug 16, 2020 at 0:25
  • 3
    $\begingroup$ If level of that course is that low then sum of this series is way beyond the scope of their understanding, the point was if it was $n^3$ in place of $n^3+1$ then the answer was known, what is the course suppose to be that can not have complex methods or zeta function? this question is way too hard for an elementary course on series. $\endgroup$
    – jimjim
    Commented Aug 16, 2020 at 0:29
  • 2
    $\begingroup$ As @jimjim suggests, $ \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3}}}}=-\frac34 \zeta(3)$, and you would not be prepared to use that. All you would have is a numerical approximation for your series of about $-0.4143011380502$ $\endgroup$
    – Henry
    Commented Aug 16, 2020 at 2:02
  • 3
    $\begingroup$ If it is a course you are developing without the zeta function it is not an appropriate problem. $\endgroup$ Commented Aug 16, 2020 at 2:30

1 Answer 1


If you do not want to use complex numbers at all, the only possibility I see is to write $$\sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=\sum\limits_{n = 1}^{p-1} {{{{{( - 1)}^n}} \over {{n^3} + 1}}}+\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}$$ For the second summation, assumming that $p$ will be sufficiently large, use $$\frac 1{n^3 + 1}=\sum_{k=1}^\infty (-1)^{k-1} \frac 1{n^{3k}}$$ and the second summation will write $$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=\sum\limits_{k = 1}^\infty(-1)^{k+1} \zeta (3 k,p)$$ which has an explicit expression in terms of polylogarithms with complex arguments !!

So, avoid the infinte upper bound for the second sum and make it as $$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\frac{\psi ^{(2)}(p)}{2}-\frac{\psi ^{(5)}(p)}{120}-\frac{\psi ^{(8)}(p)}{40320}-\frac{\psi ^{(11)}(p)}{39916800}- \cdots$$ Trying for a faw values of $p$, the decimal representaion would be $$\left( \begin{array}{cc} p & \text{approximation} \\ 25 & -0.41349860 \\ 50 & -0.41409322 \\ 75 & -0.41421222 \\ 100 & -0.41425014 \\ 125 & -0.41426913 \\ 150 & -0.41427862 \\ 175 & -0.41428481 \\ 200 & -0.41428851 \end{array} \right)$$ while the exact value would be $-0.41430114$.


If you want to go further, you can use the more general $$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\sum_{k=1}^\infty \frac{\psi ^{(3k-1)}(p)}{(3k-1)!}$$


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