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Good evening! Recently I had to study the properties of a function $$f(x):=\sum\limits_{n=1}^\infty\displaystyle\frac{x}{x^2+n^2}.$$ I found its supremum and proved that $\lim\limits_{x\to+\infty}f(x)=\displaystyle\frac{\pi}{2}$. But as far as I remember such sums can be calculated explicitly. Maybe someone knows a quick hint to this? And in general: a lot of sums can be calculated explicitly by reducing to some Taylor and Laurent series, special functions, etc., and the methods are usually special for every particular problem. Is there any book, where a complete list of known examples can be found?

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    $\begingroup$ This series is analogous to the one found here math.stackexchange.com/questions/736860/… which can be solved using the residue theorem. $\endgroup$ – JessicaK Oct 18 '14 at 0:37
  • $\begingroup$ The "bible" for such things is Gradshteyn and Ryzhik. $\endgroup$ – André Nicolas Oct 18 '14 at 1:11
  • $\begingroup$ Thanks for the link, Jessica! This is what I was looking for, to find a function with residues that equals this sum. I saw it somewhere a long time ago and forgot, of course. Quite a weird trick with ctg(z), isn't it?.) $\endgroup$ – Alexander Oct 18 '14 at 3:11
  • $\begingroup$ Gradshteyn and Ryzhik is surely a very nice book! But there ae no proofs, and as I briefly checked, doesn't contain much on function series. $\endgroup$ – Alexander Oct 18 '14 at 3:14
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Let me supplement another (simple) method for this particular problem( not the general summation technique the OP is looking for). It simply uses the Poisson's summation formula, \begin{equation} \sum_{n= -\infty}^{\infty} f(n) = \sum_{m = -\infty}^{\infty} \tilde{f}( 2\pi m ) \end{equation}

Set $g(x) = e^{-\epsilon |x| + ix \theta } $, then the Fourier transform of $g(x)$ is \begin{equation} \tilde{g}(k) =\int_{-\infty}^{\infty} e^{-\epsilon |x| + ix \theta } e^{ik x} dx = \frac{2 \epsilon}{ \epsilon^2 + (k - \theta)^2 } \end{equation} so apply Poisson's summation formula to $g(x)$, \begin{equation} \sum_{n = -\infty}^{\infty} g(n ) = \frac{\sinh \epsilon}{\cosh \epsilon - \cos \theta} = \sum_{n = -\infty}^{\infty} \frac{2 \epsilon}{ \epsilon^2 + (2\pi n - \theta )^2 } \end{equation} For the present case, set $\theta = 0$, $\epsilon = 2\pi x$, \begin{equation} RHS = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{x^2 + n^2} + \frac{1}{\pi x }= LHS = \coth( \pi x ) \end{equation} hence, \begin{equation} \sum_{n=1}^{\infty} \frac{x}{x^2 + n^2} = \frac{\pi}{2}\coth( \pi x) - \frac{1}{2x } \end{equation}

Edit: the original motivation to add the $\theta$ term is prove that following identity, \begin{equation} \frac{1}{2\pi }\frac{1- r^2}{1 -2r \cos \theta + r^2} = - \frac{1}{\pi} \sum_{-\infty}^{\infty} \frac{\ln r}{ (\ln r)^2 + ( \theta + 2\pi n )^2 } \end{equation} namely, the Poisson kernel is equal to the sum of Lorentzian(this is how I know this sum has closed form). For the present case, one can of course set $\theta = 0$ in the first place to make it "complex-analysis-free".

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  • $\begingroup$ Thanks for the idea! This method becomes "complex-analysis-free" when you set $\theta = 0$ and works nicely for $x>0$. $\endgroup$ – Alexander Oct 18 '14 at 15:55
  • $\begingroup$ I briefly checked Courant, Hilbert: they require $g$ to be $C^1$ for this method, and here $g^\prime(0)$ does not exist. But this one-point issue doesn't seem so critical, I'll think about it later... $\endgroup$ – Alexander Oct 18 '14 at 22:21
  • $\begingroup$ No need to think about it much: can be found in T.Apostol's Mathematical analysis. $\endgroup$ – Alexander Oct 21 '14 at 15:52
  • $\begingroup$ The reference pointed out by OP is Example 2(Partial fraction decomposition of $coth(x)$) in Chapter 11 section 22 of T.Apostol's Mathematical analysis. Theorem 11.24 there is the (sufficient) condition for the applicability of Poisson's summation formula. $\endgroup$ – anecdote Oct 21 '14 at 20:07
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Inspired by what JessicaK gave in her answer to a previous question (I refer to the link JessicaK gives in her comment) and using a CAS for control, what was found is $$f(x)=\sum\limits_{n=1}^\infty\displaystyle\frac{x}{x^2+n^2}=\frac{\pi x \coth (\pi x)-1}{2 x}$$ All the credit must be given to JessicaK.

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Please look at complex analysis if you want powerful and elegant methods for infinite sums such as these.

Otherwise there is also this plug 'n chug formula called the (simplified) Euler-Maclaurin formula: $$\sum_{n=1}^mf(n)=\int_1^mf(x)\textrm{d}x+\frac{1}{2}(f(1)+f(m))+\int_1^mH(x)f'(x)\textrm{d}x$$ where $f$ is differentiable and $H(x):=x-\lfloor x\rfloor-\frac{1}{2}$.

With $f(n):=x/(x^2+n^2)$ we have

$$\int_1^m \frac{x}{x^2+y^2}\textrm{d}y=\arctan(m/x)-\arctan(1/x)$$ for $m\rightarrow \infty$ this becomes $\pi/2-\arctan(1/x)$. Also $\lim_{m\rightarrow \infty}f(m)=0$.

Now for example if $x=1$ then if we neglect the messy last integral in the formula then the approximation $$\sum_{n=1}^\infty \frac{x}{x^2+n^2}\approx \frac{\pi}{2}-\arctan(1/x)+\frac{x}{2x^2+2}$$is very crude. The apprximation is $1.0353981$ for $x=1$ which is quite far from the $1/2(\pi\coth(\pi)-1)\approx 1.0766740$ given in the link provided by JessicaK. However with large $x$ the approximation is serviceable. For example for $x=1000$ i get $1.570296326$ which is pretty close to $\pi/2=1.5707963267$ which is the value that you proved convergence to as $x\rightarrow \infty$.

This is how the approximation (green) looks vs. the series (blue) for $0.5\leq x\leq 10$ enter image description here

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  • $\begingroup$ The above approximation is exactly what I was using! I obtained it by simple improper integration, and it works quite well, beacause the supremum is attained at $+\infty$... $\endgroup$ – Alexander Oct 18 '14 at 3:21
  • $\begingroup$ A nice picture... By the way, Which program is the best one for drawing such plots? Mathlab? $\endgroup$ – Alexander Oct 18 '14 at 15:09
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Maybe someone knows a quick hint to this?

Take the natural logarithm of Euler's infinite product formula for the sine function, and then differentiate it. For the final touch, use Euler's formula, linking trigonometric and hyperbolic functions together.

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