Find $\sum_{-\infty}^{\infty}\frac{1}{n^3+2} $ using complex methods.

This last one on my massive homework assignment is a real doozie! I'm going to have to finish this in the morning, but I'm having trouble assessing whether this approach (borrowed from Bak and Newman) will work. My setup is to evaluate $$\int_{C_N} \frac{\pi cot (\pi z)}{z^3+2}=\sum_{n=-N}^Nf(n)+\sum_kRes\left(\frac{\pi cot(\pi z)}{z^3+2},z_k\right)$$ where the $z_k$ are the (three simple) poles of $1/(z^3+2)$, none of which is an integer and thus of no concern for the left hand summation. With the usual "boxes" $C_N$ as indicated in Bak and Newman p. 152 (3rd Edition) and taking limits we have that $$\sum_{-\infty}^{\infty}\frac{1}{n^3+2}=-\sum_kRes\left(\frac{\pi cot(\pi z)}{z^3+2},z_k\right).$$

which should be tedius but straightforward to calculate. I am not, however, at all confident in this answer, as I am merely mimicking a similar result in another text than ours without a clear understanding of whether my procedure above is likewise valid.

  • $\begingroup$ For those of us who don't have Bak and Newman handy, could you tell us what the "boxes" are? $\endgroup$
    – David
    Mar 14, 2014 at 5:30
  • $\begingroup$ $C_N$ is a box with the top side at $Im(z) = i$, the bottom side at $Im(z) = -i$, left side at $Re(z) = -N -1/2$, and right side at $Re(z) = N+1/2$. Then $N$ is taken to infinity, and the height of the box is also taken to infinity so the contour integral becomes zero because the integrand falls of quickly with $|z|$. $\endgroup$ Mar 14, 2014 at 16:01

1 Answer 1


This approach works. First you decompose $\dfrac{1}{z^3+2}$ into partial fractions: $$\Sigma_k \frac{z_k/6}{z-z_k}.$$ Then you just do the residues one by one. You wind up getting $$\frac{1}{6}\Sigma_k \pi z_k \cot (\pi z_k).$$

I evaluated the difference between this expression and the infinite sum in mathematica, and it gave $4.4 *10^{-16}$, so I think the answer is right, but mathematica was not able to simplify the difference to $0$.


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