TITLE: How to prove that $\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^{n+1}n}{n^{2}+x^{2}}$ is always positive for all real $x$?
QUESTION [8 upvotes]: How to prove that $$\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^{n+1}n}{n^{2}+x^{2}}>0$$ for all real x?
It may looks like a very easy problem, but as it is an alternating series and does NOT converge ABSOLUTELY, it seems very hard -- or even impossible -- to use any inequalities to estimate.
Moreover, the limit $$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left ( -1 \right )^{n+1}nx^{2}}{n^{2}+x^{2}}=\frac{1}{4}$$.
If we can prove this, it will be easier to prove that the series is positive.  
I used Abel transformation formula (summation by parts) and got that it is the same value as $$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}+x^{2}}$$, and then I don't know if the problem get easier.
Now we have to prove that $\displaystyle\left|\sum_{k=n}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}+x^{2}}\right|$ is strictly decreasing for $n\in\mathbb{N}$, that is equivalent to prove that $$\frac{1}{n^2+x^2}\gt2\left|\sum_{k=n+1}^{\infty}\frac{\left(-1\right)^k}{k^2+x^2}\right|\quad \forall n\in\mathbb{N},x\in\mathbb{R}$$
Anyone has some ideas to contnue my proof?

REPLY [5 votes]: (A) : $\enspace$ As the OP Antimonius has mentioned, we can use after integration (one time) by parts 
$\hspace{1.1cm}$ the $2^{nd}$ mean value theorem for integrals. We get for  $\,\cos(x\xi)\ne 1\,$ :
$$ \int\limits_{ \frac{2\pi}{x}n }^{\frac{2\pi}{x}(n+1)} \frac{\cos(x t)}{e^t+1}dt = \frac{ ( e^{\frac{2\pi}{x}}-1)(e^{\frac{2\pi}{x}(n+1)} -1)}{( e^{\frac{2\pi}{x}n}+1)^2(e^{\frac{2\pi}{x}(n+1)}+1)^2}(1-\cos(x\xi)) > 0$$
(B) : $\enspace$ The following proof is perhaps not the shortest one but it works (without any theorems). 
Because of $\,x^2=(-x)^2\,$ it's enough to proof for $\,x\geq 0\,$ .
It's $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}n}{n^2+x^2}= \int\limits_0^\infty\frac{\cos(x t)}{e^t+1}dt \enspace $ which has for $\,x=0\,$ the result $\,\ln 2>0\,$ ,
and for $\,x> 0\,$ it’s possible to write $\enspace\displaystyle \sum\limits_{n=0}^\infty  \int\limits_{ \frac{2\pi}{x}n }^{\frac{2\pi}{x}(n+1)} \frac{\cos(x t)}{e^t+1}dt \enspace $ .
We have to show:  $\enspace\displaystyle \int\limits_{ \frac{2\pi}{x}n }^{\frac{2\pi}{x}(n+1)} \frac{\cos(x t)}{e^t+1}dt > 0 $
$\displaystyle \int\limits_{ \frac{2\pi}{x}n }^{\frac{2\pi}{x}(n+1)} \frac{\cos(x t)}{e^t+1}dt = f_n(x) + g_n(x) $
with $\enspace\displaystyle f_n(x):= \int\limits_{ \frac{\pi}{x}2n }^{\frac{\pi}{x}(2n+\frac{1}{2})} \left( \frac{1}{e^t+1} - \frac{1}{e^{t+\frac{\pi}{2x}}+1} - \frac{1}{e^{t+\frac{\pi}{x}}+1}+ \frac{1}{e^{t+\frac{3\pi}{2x}}+1} \right) \sin(x t) dt $  
and $\enspace\displaystyle g_n(x):= \int\limits_{ \frac{\pi}{x}2n }^{ \frac{\pi}{x}(2n+\frac{1}{2}) } \left( \frac{1}{e^t+1} - \frac{1}{e^{t+\frac{\pi}{x}}+1} \right) (\cos(x t) - \sin(x t)) dt$
Because of $\enspace\displaystyle \frac{1}{ac+1} - \frac{c}{ab+c} - \frac{1}{ab^2c+1} + \frac{c}{ab^3+c}=$ 
$\hspace{2.2cm}\displaystyle = \frac{ac(a^2b^3-1)(b-1)(b-c^2)(b+1)}{(ac+1)(ab+c)(ab^2c+1)(ab^3+c)} > 0 $ 
for $\,a>1\,$ , $\,b>1\,$ , $\,c>0\,$ and $\,b-c^2 > 0\,$ 
we set $\,a:=e^t>1\,$ , $\,\displaystyle b:= e^{\frac{\pi}{2x}}>1\,$ and $\,c:=1>0\,$  and get 
$\displaystyle \frac{1}{e^t+1} - \frac{1}{e^{t+\frac{\pi}{2x}}+1} - \frac{1}{e^{t+\frac{\pi}{x}}+1}+ \frac{1}{e^{t+\frac{3\pi}{2x}}+1} > 0 \,$ . 
Together with $\,\sin(xt)\geq 0\,$ for $\,\displaystyle t\in [\frac{\pi}{x}2n, \frac{\pi}{x}(2n+\frac{1}{2})]\,$ follows $\,f_n(x)>0 \,$ . 
$\displaystyle g_n(x) = \hspace{0.5cm}\sqrt{2}  \int\limits_{ \frac{\pi}{x}2n }^{ \frac{\pi}{x}(2n+\frac{1}{4}) } \left(\frac{1}{e^t+1} - \frac{1}{e^{t+\frac{\pi}{x}}+1} \right) \cos(xt+\frac{\pi}{4}) dt $ 
$\hspace{2cm}\displaystyle - \sqrt{2}\int\limits_{ \frac{\pi}{x}2n }^{ \frac{\pi}{x}(2n+\frac{1}{4}) } \left( \frac{1}{e^{t+\frac{\pi}{4x}}+1} - \frac{1}{e^{t+\frac{5\pi}{4x}}+1} \right) \sin(xt) dt  $  
Be $\displaystyle \,\delta\in [0,\frac{\pi}{4x}]\,$ . And we have $\displaystyle \,t\in [ \frac{\pi}{x}2n , \frac{\pi}{x}(2n+\frac{1}{4}) ]\,$ .
Now we compare every value of $\enspace\displaystyle \left(\frac{1}{e^t+1} - \frac{1}{e^{t+\frac{\pi}{x}}+1} \right) \cos(xt+\frac{\pi}{4}) \enspace$ with $\,t:=t_0\,$ of the given value range with $\enspace\displaystyle \left( \frac{1}{e^{t+\frac{\pi}{4x}}+1} - \frac{1}{e^{t+\frac{5\pi}{4x}}+1} \right) \sin(xt) \enspace$ with $\,t:= \frac{\pi}{x}(2n+\frac{1}{4}) -t_0 \,$ as follows:
$\displaystyle \left(\frac{1}{e^{ \frac{\pi}{x}2n +\delta } +1} - \frac{1}{e^{\frac{\pi}{x}2n +\delta +\frac{\pi}{x}}+1} \right) \cos(x(\frac{\pi}{x}2n +\delta)+\frac{\pi}{4})$
$\displaystyle >  \left( \frac{1}{e^{ \frac{\pi}{x}(2n+\frac{1}{4} ) -\delta +\frac{\pi}{4x}}+1} - \frac{1}{e^{\frac{\pi}{x}(2n+\frac{1}{4} ) -\delta +\frac{5\pi}{4x}}+1} \right) \sin(x(\frac{\pi}{x}(2n+\frac{1}{4} ) -\delta)) $
With $\displaystyle\enspace \cos(x\delta+\frac{\pi}{4})= \sin(\frac{\pi}{4}- x\delta)> 0\enspace$ we get the inequality  
$\displaystyle \frac{1}{e^{\frac{\pi}{x}2n +\delta}+1} - \frac{1}{e^{\frac{\pi}{x}2n -\delta +\frac{\pi}{2x}}+1} - \frac{1}{e^{\frac{\pi}{x}2n +\delta +\frac{\pi}{x}}+1}+ \frac{1}{e^{\frac{\pi}{x}2n -\delta +\frac{3\pi}{2x}}+1} > 0 \,$ . 
Now we can set $\displaystyle \,a:=e^{\frac{\pi}{x}2n}>1\,$ , $\displaystyle \,b:=e^{\frac{\pi}{2x}}>1\,$ , $\,c:=e^{\delta}\geq 1>0\,$ .
Because of $\,b-c^2>0\,$ for $\displaystyle \,0\leq\delta <\frac{\pi}{4x}\,$ follows, that the previous inequality is correct 
and therefore $\,g_n(x)>0\,$. With $\,f_n(x)>0\,$ and $\,g_n(x)>0\,$ the claim is proofed.