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How would we show that $$\sum_{n=1}^{\infty} \frac{1}{n^2+4n+1} = \frac{1}{6}(-2 -\pi \sqrt3 \cot(\pi \sqrt3))$$

I'm not sure how I'd go about it, but got the answer off wolfram alpha. Could someone lead me in the right direction?-series

robjohn
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    I'm unsure if what I'm about to suggest will bear fruit, it's just ideas that may or may not work. The firs thing that I would try would be partial fraction decomposition, finding two fractions, each of which have a linear polynomial in their denominator, whose sum is the pertinent fraction. ...see next comment – user2661923 Sep 10 '20 at 05:45
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    Next, assuming the partial fraction decomp worked, I would try to arrange the terms to set up a telescoping series. I suspect that this will fail here, because of the Wolfram Alpha result that you gave. ...see next comment – user2661923 Sep 10 '20 at 05:46
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    Last, I would try to express the summation as a summation of integrals with specially picked endpoints of integration. Then, I would try to use the dominant convergence theorem (50 years ago, this was referred to as a consequence of uniform continuity) to re-express everything by converting the sum of an infinite group of integrals into an integral of an infinite group of sums. Hopefully, you could then integrate each sum, and look at the limit of these, as the # of terms goes to infinity. – user2661923 Sep 10 '20 at 05:50
  • use partial fractions, and then convert that into two sums for Extended Harmonic Numbers then apply the reflection formula for Harmonic Numbers, see $(5)$ in this answer. You will need to use the recurrence relation for Harmonic Numbers, $H_n=H_{n-1}+\frac1n$, before you can use the reflection formula. Attempt this in your question, and if you can't make it work, I will post my answer. – robjohn Sep 10 '20 at 06:11
  • If the attempt I suggest above is added, it might stop the close votes. – robjohn Sep 10 '20 at 06:24

2 Answers2

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The presence of the cotangent function together with the aspect of the general term of the given series suggest relations with the eulerian development of the cotangent:

for any complex number $z \in \mathbb{C}\setminus \pi\mathbb{Z}$ one has the series development: $$\mathrm{ctg}z=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-\pi^2n^2}.$$

To begin with, since $n^2+4n+1=(n+2)^2-3$ the given series can be rewritten as: $$\sum_{n=1}^{\infty}\frac{1}{n^2+4n+1}=\sum_{n=3}^{\infty}\frac{1}{n^2-3},$$ through a very clear bijective change of summation variable. Let us denote the sum of the series in question by $s$.

Taking $z=\pi\sqrt{3}$ in the forementioned development we obtain: $$\mathrm{ctg}(\pi\sqrt{3})=\frac{1}{\pi\sqrt{3}}+\frac{2\sqrt{3}}{\pi}\sum_{n=1}^{\infty}\frac{1}{3-n^2}=\frac{1}{\pi\sqrt{3}}+\frac{2\sqrt{3}}{\pi}\left(\frac{1}{2}+\frac{1}{-1}-s\right)$$ whence after an elementary rearrangement we gather: $$\sum_{n=1}^{\infty}\frac{1}{n^2+4n+1}=-\frac{1}{6}\left(2+\pi\sqrt{3}\mathrm{ctg}\left(\pi\sqrt{3}\right)\right).$$

ΑΘΩ
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$$S=\sum_{1}^{\infty} \frac{1}{n^2+4n+1}= \frac{1}{b-a} \left(\sum_{n=0}^{\infty} \frac{1}{n+a}-\frac{1}{n+b}\right) dx.$$ Here $a=2-\sqrt{3}, b=2+\sqrt{3}.$ Next, using integral tepresentation of: $1/z=\int_{0}^{1}t^{z-1} dt$, we get $$S=(b-a)^{-1}\int_{0}^{1}\sum_{n=1}^{\infty} [t^{n+a-1}-t^{n+b-1}] dt = (b-a)^{-1}\int_{0}^{1} \frac{t^{a}-t^{b}}{1-t}dt.$$ We have used $\sum_{1}^{\infty} z^n=\frac{z}{1-z},~if~|z|<1.$ Next using the definition of Harmonic numbers we can write $$S=H_b-H_a$$ I may get back again. Can some one take it from here?

Z Ahmed
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