Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$ is not uniformly convergent on $[0,\;\frac{\pi}{2})$? Why $\sum_{n=1}^{\infty}\frac{n\left(\sin x\right)^{n}}{2+n^{2}}$
is not uniformly convergent on $[0,\;\frac{\pi}{2})$? 
I was thinking that we need to show partial sums
\begin{equation}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right|\nrightarrow0
\end{equation}
for some $x\in[0,\;\frac{\pi}{2})$. 
Any hint? 
Is this correct?
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & =\left|\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\right|\\
 & =\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(n+1\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
 & \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{2n\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
 & \geq\frac{\left(n+1\right)\left(\sin x\right)^{n+1}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
 & \geq\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}+\cdots+\frac{\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
 & =\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber 
\end{align}
then let $x=\frac{\pi}{2}-\frac{1}{n}$, so we have 
\begin{align}
\left|S_{2n}\left(x\right)-S_{n}\left(x\right)\right| & \geq\frac{n\left(n+1\right)\left(\sin x\right)^{2n}}{2+\left(2n\right)^{2}}\\
 & =\frac{n\left(n+1\right)\left(\sin\left(\frac{\pi}{2}-\frac{1}{n}\right)\right)^{2n}}{2+\left(2n\right)^{2}}\nonumber \\
 & \rightarrow0.25,\:not\:0\nonumber 
\end{align}
 A: First, your proof looks fine to me. Below I sketch a different kind of proof.
Let $f(x)$ be the sum of this series. Note that each summand increases on $[0,\pi/2).$ Hence so does $f(x).$ It follows that $\lim_{x\to \pi/2^-} f(x)=L$ for some $L\in (0,\infty].$
Suppose the series converges uniformly to $f$ on $[0,\pi/2).$ Because each summand is bounded on $[0,\pi/2),$ so is every partial sum, which implies by uniform convergence that $f$ is bounded on $[0,\pi/2).$ Thus $L<\infty.$
Let $N\in \mathbb N.$ Then
$$\tag 1 L =  \lim_{x\to \pi/2^-} f(x) \ge \lim_{x\to \pi/2^-}\sum_{n=1}^{N}\frac{n(\sin x)^{n}}{2+n^{2}} = \sum_{n=1}^{N}\frac{n}{2+n^{2}}.$$
Since $\sum_{n=1}^{\infty}\dfrac{n}{2+n^{2}} = \infty,$ the right side of $(1)$ can be made arbitrarily large. Thus $L=\infty,$ contradiction.
A: For any functions $f,g$ from $[0,\pi/2)$ to $\Bbb R$ let $\|f-g\|=\sup  \{|f(x)-g(x)|: x\in [0,\pi/2)\}\in [0,\infty].$
For $m\in \Bbb N$ let $S_m(x)=\sum_{n=1}^m(n(\sin x)^n)/(2+n^2).$
Suppose that $S_m\to S$ uniformly on $[0,\pi/2).$ Then  $\|S-S_m\|\to 0.$ Then $\lim_{M\to \infty}\sup_{M<m<m'}\|S_m-S_{m'}\|=0$ because $\|S_m-S_{m'}\|\leq \|S_m-S\|+\|S-S_{m'}\|.$
Given $M\in \Bbb N,$ take $m>M$ and let $m'=2m+1.$ Then take $x\in [0,\pi/2)$ where $x$ is close enough to $\pi /2$ that $(\sin x)^{2m+1}>1/2.$ Then $(\sin x)^j>1/2$ for $m+1\leq j\leq 2m+1.$ And we have  $$\|S_m-S_{m'}\|\geq  |S_m(x)-S_{m'}(x)|=$$ $$=\sum_{j=m+1}^{2m+1}\frac {j(\sin x)^j}{j^2+2}> 
 \sum_{j=m+1}^{2m+1}\frac {j/2}{j^2+2}>$$ $$> \sum_{j=m+1}^{2m+1}\frac {j/2}{2j^2}=\sum_{j=m+1}^{2m+1} \frac {1}{4j}>$$ $$> \frac {1}{4}\sum_{j=m+1}^{2m+1}\int_j^{j+1}(\frac {1}{x})dx=$$  $$=\frac {1}{4}\int_{m+1}^{2m+2}(\frac {1}{x})dx=\frac {1}{4}\ln 2.$$
So convergence of $S_m$ cannot be uniform.
