Show that the following series is not uniformly convergent: $$\sum_{n=1}^{\infty}{\dfrac{n}{2+n^2}\left(\sin(x)\right)^n} \quad
\text{in} \quad [ 0,\frac{\pi}{2})$$
Attempt: One idea I thought of, is to show that the function $$f(x) = \sum_{n=1}^{\infty}{f_n(x)}$$ is not continuous. This is apparent using $x=0$ and $x=\pi/2$ (amongst the few values for which it fails to be continuous).
My question is if this is sufficient? Or would I have to find the sum function for all values of $x$?
Also, we know that $$\left|\dfrac{n}{2+n^2}\left(\sin(x)\right)^n\right| < M_n =\dfrac{n}{2+n^2}$$ and since $\sum{M_n}$ diverges the given series is not uniformly convergent. Is this reasoning valid?
Best Answer
Your converse $M$-test argument is not valid $\dots$ if it were it would show that the zero series is not uniformly convergent.
The sum of a uniformly convergent series must be bounded but it's pretty obvious the sum of your series is not bounded on $[0, \pi/2)$ since $\sum\frac{n}{2+n^2} = \infty$, the $\sin$ can be made arbitrarily close to $1$ and everything in sight is non-negative.