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.


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 $1/ \sqrt{2} \leqslant \sin n x_m \leqslant 1$.


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.