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For $\alpha\in\mathbb{R}$, it seems that we have $\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi}{\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right).$

I've tried many techniques and ways -- partial fractions, turning it into an integral, finding the partial sum and so on -- to show it, but all ended in failure. Many thanks for the help in advance.

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    Smells like a residue computation. Are you familiar with complex analysis?2011-02-21
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    Hint: use $\tan(z)=\sum_{n \; odd} 1/(\frac{n\pi}{2} - z)$ with the sum over all odd integers (negative and positive). You can find this with residue techniques as Qiaochu suggests.2011-02-21
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    btw, should that be arctan instead of tan?2011-02-21
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    @Qiaochu Yuan, @Raskolnikov -- thanks for the suggestions/hints. (I'm uninitiated in complex analysis but despite that I would still be of course more than happy to receive a proof that uses complex analysis). I will see what I can do with the hints.2011-02-22
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    @Moron, no, it's tan. I've tested it with large number of infinite series of that form; e.g. take: $$\displaystyle S = \sum_{k \ge 0}\frac{1}{16k^2+16k+3} = \frac{\pi}{16\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right)\bigg|_{ \alpha = -\frac{3}{16}} = \frac{\pi}{8}$$, which writing the series as an integral, $\frac{1}{2}\int_0^{1}\frac{1}{1+x^2}\;{dx}$ agrees!2011-02-22
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    +1 @Raskolnikov: You should write out the solution so that this question gets an answer. @TCM: You need not know a lot of complex analysis. If you are fine with slightly less rigorous argument, (Euler's proof of Basel's problem was not rigorous) all you need to note is that $\tan(z)$ goes to infinity as $z \rightarrow \frac{k \pi}{2}$ where $k$ is an odd integer and that limit $(z-\frac{k \pi}{2}) \tan(z) = -1$ as $z \rightarrow \frac{k \pi}{2}$ which allows us to write $\tan(z) = \displaystyle \sum_{k=odd} \frac{1}{\frac{k \pi}{2}-z}$2011-02-22
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    @Sivaram: Derek already wrote a brilliant answer. I also think that generally, we should not always run too fast to write out an answer. Sometimes giving hints is nicer to the person asking, let them some of the pleasure of discovery.2011-02-22
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    @Raskolnikov: My comment was before Derek typed out the answer and I meant your comment/hint could be written as an answer.2011-02-22

1 Answers 1

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First we note that

$$ S = \sum_{k \ge 0} \frac{1}{k^2+k-\alpha} = \sum_{k \ge 0} \frac{1}{(k+1/2)^2 - (4\alpha+1)/4} = 4\sum_{k \ge 0} \frac{1}{(2k+1)^2 -a^2},$$

where $a=\sqrt{4\alpha+1},$ and so

$$ S = 4 \sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2}.$$

Now we use the well-known cotangent identity (at the bottom of the page)

$$\pi\cot(a\pi) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-m^2} . \quad (1)$$

Replacing $a$ by $a/2$ and dividing by $2$ gives

$$ \frac12 \pi\cot \left( \frac{a\pi}{2}\right) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-(2m)^2} . \quad (2)$$

Subtracting $(2)$ from $(1)$ gives

$$ \pi \cot(a\pi) - \frac12 \pi\cot \left( \frac{a\pi}{2}\right) = \sum_{m=1, \textrm{ odd } m}^\infty \frac{2a}{a^2-m^2} $$

but $\cot(\theta) -\frac12 \cot(\theta/2)= -\frac12 \tan(\theta/2)$ and so

$$ \frac{\pi}{4a} \tan \left( \frac{a \pi}{2} \right) = \sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2},$$

from which the result follows setting $a = \sqrt{4\alpha + 1}.$

EDIT: The cotangent identity is proven here.

EDIT2: An easy way to discover the cotangent identity is to take logarithms of the following product formula for $\sin \pi x$ and differentiate wrt $x.$ $$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right)$$

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    @Sivaram: Sorry, I only popped down to the computer because I'm not sleeping well tonight but it's back to bed now... although I'm not sure maths is the best soporific :-)2011-02-22