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Let $f(x)=\sum_{n=1}^{\infty}\frac{\sqrt n}{n+x}\sin nx$. Show that for any $x\in (-1,\infty)$ the series is convergent. Find the intervals on which the series is uniformly convergent.

If $x=k\pi$, where $k$ is an integer, then $\sin nx=0$. Now suppose $x\neq k\pi$, then $|\sum \sin kx|\leq \frac{1}{\sin{\frac{x}{2}}}$.

As $x\in (-1,\infty)$, then $\frac{1}{n+x}<\frac{1}{n-1}$, which implies $\frac{\sqrt n}{n+x}<\frac{\sqrt n}{n-1}$. Hence, for any $x\in (-1,\infty)$ by the Dirichlet test, we can guarantee that the series is convergent at least pointwise.

But I need help for uniformly convergent part. Thank you.

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The easier part is to show that the series is uniformly convergent on any compact interval $[c_1,c_2]$ that excludes $-1$ and $0$ as well as any integer multiple of $\pi$. This follows from an application of the Dirichlet test for uniform convergence, since $\sqrt{n}/(n+x)$ converges monotonically and uniformly to $0$ for sufficiently large $n$, and as you observed $\left|\sum_{n=1}^m \sin(nx)\right|$ is uniformly bounded for all $m$ on such intervals.

Convergence is non-uniform on any interval such as $(0,c_2]$ where $0$ is a limit point. In this case the Cauchy criterion for uniform convergence is violated.

For any $m \in \mathbb{N},$ let $x_m = \pi/(4m)$. With $m \leqslant n \leqslant 2m$, we have $\pi/4 \leqslant nx_m \leqslant \pi/2$ and $1/ \sqrt{2} \leqslant \sin n x_m \leqslant 1$.

Hence,

$$\left|\sum_{n = m}^{2m} \frac{\sqrt{n}}{n + x_m}\sin (nx_m)\right| \geqslant \frac{1}{\sqrt{2}}\sum_{n=m}^{2m}\frac{\sqrt{n}}{n + \pi/4}.$$

Since the series $\sum_n \sqrt{n}/(n + \pi/4)$ diverges, the RHS cannot be arbitrarily small, regardless of the choice for $m.$

I shall leave it for you to show that convergence is not uniform on any interval with $-1$ as a limit point.

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