Determine uniform convergence of series

Question

I have problems with checking if series $\displaystyle S_n(x)= \sum _{n=1} ^{\infty} \frac{1}{n(1+(x-n)^2)}$ is uniform convergent at $(0 , +\infty)$

My try: consider $\displaystyle |S_n(x) - S(x)|=\sum _{k=n+1} ^{\infty} \frac{1}{k(1+(x-k)^2)}$, let $m=n+1$ and we have

$\displaystyle \sum _{k=m} ^{\infty} \frac{1}{k(1+x^2-2xk+k^2)}$ is series of positive terms so for $x= \sqrt{\frac{1-m}{m}}+m$ we have that first term of this series is $\displaystyle \frac{1}{m(1+\frac{1-m}{m})}=1$ so $\displaystyle |S_n(x) - S(x)|>1>\epsilon$ so it's not uniform convergent

Answer 1

Consider the sequence $b_n (x) =\frac{1}{1+ (x-n)^2} .$ Take any $x>0$ and let $\lfloor x\rfloor =m$ then $$\sum_{j=1}^{\infty} b_j (x) =\sum_{j=1}^m b_j (x) +\sum_{j=m+1}^{\infty} b_j (x) =\sum_{j=1}^m \frac{1}{1+ (x-j)^2} +\sum_{j=m+1}^{\infty} \frac{1}{1+ (x-j)^2} \leqslant 2+ \sum_{j=2}^{m-1} \frac{1}{1+ (x-j)^2} +1+\sum_{j=m+2}^{\infty} \frac{1}{1+ (x-j)^2} \leq 3 +\sum_{j=2}^{m-1} \frac{1}{1+ j^2} +\sum_{j=m+2}^{\infty} \frac{1}{1+ j^2}\leq\hspace{1cm}\leq 3+2\sum_{j=1}^{\infty} \frac{1}{1+ j^2} <\infty$$ Hece the sequence $s_n (x) =\sum_{j=1}^{n} b_j (x)$ is uniformly bounded on $(0,\infty)$. The sequence $a_n(x) =\frac{1}{n} $ is monotone and tends uniformly to $0$ therefore by Dirichlet Uniform Convergence Test the series $$\sum_{n=1}^{\infty} a_n (x) b_n (x) $$ is uniformly convergent on $(0,\infty ).$

Answer 2

The $n$-th term of the series is concentrated around $x=n$, hence $$\sum_{n=1}^{2N}\frac{1}{n(1+(x-n)^2)}-\sum_{n=1}^{N}\frac{1}{n(1+(x-n)^2)}$$ is big on the interval $x\in(N,2N)$. Not bigger than $\frac{C}{N}$, in any case, so the original series is uniformly convergent on $\mathbb{R}^+$. To check it is pointwise convergent, we may also use: $$\frac{1}{1+a^2}=\int_{0}^{+\infty}\cos(az)\,e^{-z}\,dz,$$ hence: $$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{n(1+(x-n)^2)} &=& \int_{0}^{+\infty}e^{-z}\sum_{n\geq 1}\frac{\cos((x-n)z)}{n}\,dz\\ &=&-\text{Re}\int_{0}^{+\infty}e^{-z}e^{ixz}\log(1-e^{-iz})\,dz\\&=&\text{Re}\left[\frac{\pi+i H_{x+i-1}}{x+i}\right]=\text{Re}\left[\frac{\pi+i\gamma+i\psi(x+i)}{x+i}\right]\end{eqnarray*}$$ and our limit function decays like $\frac{1}{x}$ for large $x$.