Uniform convergence of the function series $\sum_{n\geq 0}\frac{\sin{x}}{1+n^2 x^2}$

Question

Is the series $$S(x):=\sum_{n\geq 0}\frac{\sin{x}}{1+n^2 x^2}$$

uniformly convergent for small $x$ ?

Answer 1

Hint: If the series were uniformly convergent on some $[0,a],$ then for large $N,$

$$\sum_{n=N}^{2N}\frac{\sin x}{1+n^2x^2}$$

would be less than $1/10$ for $x\in [0,a].$ Try $x=1/N$ to see this doesn't happen.

Answer 2

Following the hint of zhw,

If $S_{n}(x)=\sum_{k=0}^{n}\frac{\sin{x}}{n^2 x^2+1}$ convereges uniformly to some $S(x)$ on the interval $[-1,1]$ then the tail sequence $T_{n}(x)=\sum_{k=n}^{\infty}\frac{\sin{x}}{k^2 x^2+1}$ should converge to zero for every $x\in [-1,1]$.

But for $n\sim \frac{1}{\delta}$ $$ | T_{n}(\delta)|= \sin{(\delta)}\,\sum_{n\sim \frac{1}{\delta}} \frac{1}{1+n^2 \delta^2}\sim \frac{\sin{\delta}}{\delta} \sim 1 $$

So there does not exist any $N$ however large such that the tail $T_{n}(x)$ is arbitrarily small for all (and this is key) $x$, when $n\geq N$. If such $N$ exists we would have $T_{N}(x)$ arbitrarily small for all $x\in [-1,1]$ but have just seen $T_{N}(\frac{1}{N})\gtrsim 1$.

Answer 3

Thanks @Jack D'Aurizio pointing out my error. Following the hint given by @zhw. we have \begin{align*} \sum_{n=N}^{2N}\dfrac{\sin x}{1+n^{2}x^{2}}<\dfrac{1}{2}(\tan^{-1}2-\tan^{-1}1) \end{align*} for large $N$ and $x\in[0,a]$. Putting $x=1/N$, \begin{align*} \sin(1/N)\sum_{n=N}^{2N}\dfrac{1}{1+(n/N)^{2}}&\geq\sin(1/N)\int_{N}^{2N}\dfrac{1}{1+(x/N)^{2}}dx\\ &=N\sin(1/N)\int_{1}^{2}\dfrac{1}{1+u^{2}}du\\ &=\dfrac{\sin(1/N)}{1/N}(\tan^{-1}2-\tan^{-1}1), \end{align*} taking $N\rightarrow\infty$ will violate the inequality.