Examine uniform convergence of function series $\sum_{n=1}^{+\infty} \frac{1}{n(1+(x-n)^2)}$ 
Examine uniform convergence of function series $\sum_{n=1}^{+\infty} \frac{1}{n(1+(x-n)^2)}$

My try:
Let $$g_n(x)=\frac{1}{n(1+(x-n)^2)}$$  Note that for $$f_n=\frac{\arctan (x-n)}{n}$$ we have: $$f'_n(x)=\frac{1}{n(1+(x-n)^2)}=g_n(x)$$So if $\sum g_n$ is pointwise convergence then $\sum f'_n$ is uniform convergence. However we know that for enought big $n$: $f'_n \le \frac{-1}{n}$ so number series $\sum f'_n$ is divergent so function series $\sum f'_n$ is not pointwise convergent so also is not uniform convergent. That's why $\sum g_n$ is not pointwise convergent and also uniform.However I have doubt if this fact is "$\Leftrightarrow$" or implication and it does not work in the side in which I used it.Can you check it?
 A: I'm not sure I follow your argument with $f_n'$. $f_n'$ is never negative, so the inequality $f_n'<-1/n$ can't be right. Perhaps you meant $f_n$? But you are going in the right direction - just because the $f_n$ series diverges doesn't mean the $g_n$ series does. In fact, $\sum g_n$ converges uniformly on $\mathbb{R}$. Here's a proof, but it's quite finicky in places.  
Let $G_n=\sum_{k=1}^n g_k$ and $G=\lim_{n\rightarrow\infty}G_n$ (right now we don't even know if it exists). Always keep in mind that all the terms we are dealing with are positive. The strategy is the following: suppose we manage to show that


*

*$G_n$ converges uniformly on every bounded interval,

*$\lim_{x\rightarrow\infty}G(x)=0$.


Say we've done this. Since $g_n(-x)\leq g_n(x)$, it's also true that $G(-x)\leq G(x)$ and so $\lim_{x\rightarrow\infty}G(x)=0$ implies $\lim_{x\rightarrow-\infty}G(x)=0$. Now for uniform continuity on $\mathbb{R}$ we must show that $$\forall\epsilon>0\,\exists N\in\mathbb{N}:\forall n>N\,\,\forall x\in\mathbb{R}\,\left\vert G(x)-G_n(x)\right\vert <\epsilon$$
Basically, we only need to go up to $N$ and we'll be pretty close to $G$ everywhere.
Now pick an $\epsilon$. Due to $\lim_{x\rightarrow\pm\infty}G(x)=0$ we can choose a bounded interval $I$ so big that $G < \epsilon$ everywhere outside $I$. Since $0<G_n(x)<G(x)$, automatically we have $\left\vert G(x)-G_n(x)\right\vert <\epsilon$ on $\mathbb{R}-I$. But since $I$ is bounded, we can pick an $N$ so big that for all $n>N: \left\vert G(x)-G_n(x)\right\vert <\epsilon$ holds inside $I$ as well, thus proving convergence is uniform on $(\mathbb{R}-I)\cup\mathbb{R}=\mathbb{R}$.
Now for the two statements.



*

*$G_n$ converges uniformly on every bounded interval


Let $(a,b)$ be our interval. We can pick $K$ large enough that for $k>K$ we have $k-x > k/2$ for all $x\in(a,b)$ and so for such $k$:
$$
g_k(x) = \frac{1}{k(1+(k-x)^2)}<\frac{1}{k(1+(k/2)^2)}<\frac{4}{k^3}
$$
The series $\sum k^{-3}$ is well-know to be convergent and so $\sum g_k(x)$ is convergent on $(a,b)$. Furthermore, we estimate the remainder:
$$
\vert G(x)-G_n(x)\vert = \sum_{k=n+1}^\infty g_k(x)<\sum_{k=n+1}^\infty \frac{4}{k^3}
$$
As this bound on the remainder is independent of $x$, we have uniform convergence on $(a,b)$ (notice, however, that this does not suffice for uniform convergence on $\mathbb{R}$, as $K$ depends on the choice of interval).



*

*$\lim_{x\rightarrow\infty}G(x)=0$
This is somewhat convoluted, but I didn't find a simpler way. We would like to take 
$$
G(x) = \sum_{k=1}^\infty \frac{1}{k(1+(k-x)^2)}
$$
and somehow bound it from above by something that vanishes at infinity. So, pick a big $x$ and split $G(x)$ into three parts:
$$
G(x) = S_1(x)+S_2(x)+S_3(x)\\
S_1(x)=\sum_{1\leq k<x-x^{2/3}} \frac{1}{k(1+(k-x)^2)}\\
S_2(x)=\sum_{x-x^{2/3}\leq k\leq x+x^{2/3}}\frac{1}{k(1+(k-x)^2)}\\
S_3(x)=\sum_{x+x^{2/3}<k}\frac{1}{k(1+(k-x)^2)}
$$
Let's get to bounding!


*

*$S_1$
It is true that $x-k > x^{2/3}$, so we have:
$$
S_1(x)=\sum_{1\leq k<x-x^{2/3}} \frac{1}{k(1+(k-x)^2)}<\sum_{1\leq k<x-x^{2/3}}\frac{1}{1+(x^{2/3})^2}<\frac{x}{1+x^{4/3}}<x^{-1/3}
$$ 


*

*$S_2$
$$
S_2(x)=\sum_{x-x^{2/3}\leq k\leq x+x^{2/3}}\frac{1}{k(1+(k-x)^2)}<\sum_{x-x^{2/3}\leq k\leq x+x^{2/3}}\frac{1}{x-x^{2/3}}\leq\frac{2x^{2/3}+2}{x-x^{2/3}}
$$

*$S_3$
$$
S_3(x)=\sum_{x+x^{2/3}<k}\frac{1}{k(1+(k-x)^2)}<\sum_{x+x^{2/3}<k}\frac{1}{(k-x)^3}<
\sum_{x^{2/3}-1\leq k}\frac{1}{k^3}
$$
$S_1$, $S_2$ and $S_3$ all go to $0$ at infinity ($S_3$ because it is the remainder of $\sum k^{-3}$), thus finally establishing $\lim_{x\rightarrow\infty}G(x)=0$.

Now that uniform convergence has been established, we can also integrate $G$ termwise:
$$
\int_0^x G(t)\mathrm{d}t=\sum_{k=1}^\infty\frac{\mathrm{arctan}(x-k)+\mathrm{arctan(k)}}{k}
$$
Now this isn't precisely the $f_k$ series you had, it has an extra arctangent in the numerator. Without it, the series would diverge for all $x$, as you correctly pointed out. I believe the problem was that you just took an indefinite integral of the terms, but (at least Wikipedia makes it seem that) you can only take definite integrals and you simply can't get the $f_n$ series you described out of the definite integral.
