$$S = \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3} \space \text{using complex analysis}$$

This my question: we need to consider a $f(z)$ such that,

$$\frac{1}{2\pi i} \cdot\oint_{C_N} f(z) dz = \text{something (maybe residue)} + \sum_{n=1}^{N} \frac{1}{(n+1)(n+2)^3}$$

How do we do this? I believe we consider, a square with vertices:


We also need to prove that as $N \to \infty$ that $\displaystyle \oint_{C_N} f(z) dz \to 0$

Can someone give me an idea?

  • $\begingroup$ hmmm, sounds interesting but i have unfortunately no time to get into the details today. Maybe something like $\psi^{(2)}(z)\psi^{(0)}(z)$ would be a candidate. $\psi^{(m)}(z)$ is a polygammafunction of order $m$ $\endgroup$ – tired Jan 12 '15 at 14:40
  • $\begingroup$ I guess you wanted to use something with $\pi\cot(\pi z)$, but I'm not sure that one should do so here (as in your other post on the Basel problem). After all, you will also end up with a sum over negative integers, and I see no direct symmetry... Maybe one can take care of that... $\endgroup$ – mickep Jan 12 '15 at 14:45
  • $\begingroup$ Yeah, it is a problem... What can I do? $\endgroup$ – Amad27 Jan 12 '15 at 15:19
  • $\begingroup$ maybe this reference will be of sum interest for you: projecteuclid.org/download/pdf_1/euclid.em/1047674270 $\endgroup$ – tired Jan 12 '15 at 15:29

Here is an unorthodox answer that needs some more work to make it rigorous. Consider $$f(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}.$$ We can certainly integrate around a circular contour as ML gives $2\pi R\times \log R/R^4.$

Now we have $$\mathrm{Res}_{z=0} f(z) = \frac{1}{8},$$ $$\mathrm{Res}_{z=-1} f(z) = -\gamma,$$ $$\mathrm{Res}_{z=-2} f(z) = \frac{\pi^2}{6} +\zeta(3) - 3 +\gamma,$$ and finally $$\mathrm{Res}_{z=n} f(z) = \frac{1}{(n+1)(n+2)^3},$$ This means that the sum is $$\frac{23}{8} - \frac{\pi^2}{6} -\zeta(3).$$

Addendum. To see how to calculate the residue of $f(z)$ at $z=-2$ use the Cauchy Integral Formula for the series coefficients which is $$\frac{1}{2\pi i} \int_{|z+2|=\epsilon} \frac{\psi(-z)}{(z+2)^{n+1}} \; dz.$$

Put $z+2 = -w$ to get $$- \frac{(-1)^{n+1}}{2\pi i} \int_{|w|=\epsilon} \frac{\psi(w+2)}{w^{n+1}} \; dw$$

which is $$\frac{(-1)^n}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \left(\frac{1}{w+1} + \psi(w+1)\right) \; dw$$

The first component can be evaluated to give the series $$1+(z+2)+(z+2)^2+(z+2)^3+\cdots$$ (remember that the CIF integral gives the coefficient on $(z+2)^n.$)

The second component is $$\frac{(-1)^n}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \psi(w+1) \; dw$$

The rational zeta series for the digamma function is $$\psi(z+1) = -\gamma - \sum_{k\ge 1} \zeta(k+1) (-z)^k$$ for $|z|<1.$

It follows that the second component gives the series $$-\gamma - \sum_{k\ge 1} \zeta(k+1) (z+2)^k.$$

This finally yields $$\psi(-z) = 1-\gamma + \sum_{k\ge 1} (1-\zeta(k+1)) (z+2)^k.$$

Now since $$\frac{1}{z+1} = \frac{1}{-1+z+2} = -\frac{1}{1-(z+2)} = -1 - (z+2) - (z+2)^2 - (z+2)^3 -\cdots$$

The desired residue is $$-(1-\zeta(3)) - (1-\zeta(2)) - (1-\gamma) = -3 + \gamma + \frac{\pi^2}{6} + \zeta(3).$$

  • $\begingroup$ This is a great answer, I do have a few questions. Firstly, how did you compute the residue at $z=n$ for any $n$? Also arent you consider $-n$? or $\pm ni$? $\endgroup$ – Amad27 Jan 13 '15 at 12:20
  • $\begingroup$ Also, I dont understand how you found the coefficient at $z=-2$ $\endgroup$ – Amad27 Jan 13 '15 at 12:30

This approach will not get you the sum you want. Let's see why.

The contour $C_N$ you show encloses simple poles at all of the integers except at $z=-1$, which has a double pole, and $z=-2$, which has a quadruple pole. I will not compute them here as I do not need to in order to make my point.

We may safely assume that the contour integral

$$\oint_{C_N} dz \frac{\pi \cot{\pi z}}{(z+1)(z+2)^3} $$

vanishes as $N \to \infty$. By the residue theorem, however, this contour integral is $i 2 \pi$ times the sum of the residues of the poles inside $C_N$. Thus, we have

$$\sum_{n=-\infty}^{-3} \frac1{(n+1)(n+2)^3} + \sum_{n=0}^{\infty} \frac1{(n+1)(n+2)^3} + \left [\frac{d}{dz} \frac{\pi \cot{\pi z}}{(z+2)^3} \right ]_{z=-1} + \frac1{3!}\left [\frac{d^3}{dz^3} \frac{\pi \cot{\pi z}}{z+1} \right ]_{z=-2}= 0$$

Note that the first sum does not match up with the second sum. It is easier to see this after rewriting the first sum by subbing $n \mapsto -n$ to get

$$\sum_{n=3}^{\infty} \frac1{(n-1) (n-2)^3} = \sum_{n=0}^{\infty} \frac1{(n+1)^3 (n+2)} $$

Thus, we find that the sum of two sums is equal to the residues at the poles...blah blah blah. But not the sum you seek.

  • $\begingroup$ So this complex analysis method doesnt work with odd/not odd nor even sums? Thats a bummer. $\endgroup$ – Amad27 Jan 12 '15 at 16:06
  • $\begingroup$ @Amad27: I would hesitate to make such a sweeping statement like that. Maybe you could find a different $f$ with a different set of poles. Do some research. But the standard way of evaluating sums using residues won't work here - the $\zeta(3)$ in the answer should be a clue. $\endgroup$ – Ron Gordon Jan 12 '15 at 16:07

This is just a long comment. What does it mean to evaluate? Observe that $$\frac{1}{(n+1)(n+2)^3}=(\frac{1}{n+1}-\frac{1}{n+2})\frac{1}{(n+2)^2}$$ $$=-\frac{1}{(n+2)^3}+\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)^2}$$ So your answer will in one way or another involve $\zeta(3)$. I guess what you want is just an interpretation of the sum as a contour integral.

  • $\begingroup$ Also, $1/((n+1)(n+2))=1/(n+1)-1/(n+2)$, so that part is telescoping. But then, the complete sum is calculated without residue calculus :) $\endgroup$ – mickep Jan 12 '15 at 14:55

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