How to calculate the sum of $\sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}} $ 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.
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
$$\left(
\begin{array}{cc}
 p &amp; \text{approximation} \\
 25 &amp; -0.41349860 \\
 50 &amp; -0.41409322 \\
 75 &amp; -0.41421222 \\
 100 &amp; -0.41425014 \\
 125 &amp; -0.41426913 \\
 150 &amp; -0.41427862 \\
 175 &amp; -0.41428481 \\
 200 &amp; -0.41428851 
\end{array}
\right)$$ while the exact value would be $-0.41430114$.
Edit
If you want to go further, you can use the more general
$$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\sum_{k=1}^\infty \frac{\psi ^{(3k-1)}(p)}{(3k-1)!}$$
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
$$\left(
\begin{array}{cc}
 p &amp; \text{approximation} \\
 25 &amp; -0.41349860 \\
 50 &amp; -0.41409322 \\
 75 &amp; -0.41421222 \\
 100 &amp; -0.41425014 \\
 125 &amp; -0.41426913 \\
 150 &amp; -0.41427862 \\
 175 &amp; -0.41428481 \\
 200 &amp; -0.41428851 
\end{array}
\right)$$ while the exact value would be $-0.41430114$.
Edit
If you want to go further, you can use the more general
$$\sum\limits_{n = p}^\infty {{{{{( - 1)}^n}} \over {{n^3} + 1}}}=-\sum_{k=1}^\infty \frac{\psi ^{(3k-1)}(p)}{(3k-1)!}$$