Prove that f(x) = the sum from n = 2 to infinity of sin(nx)/(n ln n) is continuous at x = 0.

Step-by-step solution

1. 1

   Reduce to a limit statement, and note why the easy tools fail. Every term vanishes at $x=0$, so $f(0)=0$ and continuity means $f(x)\to0$ as $x\to0$. The Weierstrass $M$-test is useless here: $\sum 1/(n\ln n)$ diverges. Dirichlet's test gives uniform convergence only on $[\delta,2\pi-\delta]$ — precisely away from the point of interest. So the sum must be split by hand. Since $f$ is odd, it suffices to treat $0<x<\tfrac12$.

2. 2

   Choose the splitting point. Let $N=\lfloor 1/x\rfloor$, so that $N\le 1/x<N+1$ and $N\to\infty$ as $x\to0^{+}$. Write

   $f(x)=\underbrace{\sum_{n=2}^{N}\frac{\sin nx}{n\ln n}}_{S_1}+\underbrace{\sum_{n=N+1}^{\infty}\frac{\sin nx}{n\ln n}}_{S_2}.$

   For $n\le N$ the angle $nx\le 1$ is small, so $\sin nx$ can be bounded crudely; for $n>N$ the sine oscillates and cancellation must be exploited instead. Two regimes, two different estimates.

3. 3

   Bound the head $S_1$ with $|\sin nx|\le nx$.

   $|S_1|\le\sum_{n=2}^{N}\frac{nx}{n\ln n}=x\sum_{n=2}^{N}\frac{1}{\ln n}.$

   Splitting that sum at $\sqrt N$ — below it use $1/\ln n\le 1/\ln 2$, above it use $1/\ln n\le 2/\ln N$ — gives

   $x\sum_{n=2}^{N}\frac1{\ln n}\le \frac{x\sqrt N}{\ln 2}+\frac{2xN}{\ln N}\le\frac{1}{\sqrt N\,\ln 2}+\frac{2}{\ln N},$

   using $xN\le 1$ and $x\sqrt N\le 1/\sqrt N$. Both pieces tend to $0$. The naive bound $1/\ln n\le1/\ln 2$ alone would only give $O(1)$, which is why the split at $\sqrt N$ is needed.

4. 4

   Bound the tail $S_2$ by Abel summation. The Dirichlet kernel estimate gives, for $0<x\le\pi$ and any $q>p$,

   $\left|\sum_{n=p}^{q}\sin nx\right|\le\frac{1}{\sin(x/2)}\le\frac{\pi}{x},$

   since $\sin(x/2)\ge x/\pi$ on $(0,\pi]$. As $a_n=1/(n\ln n)$ decreases to $0$, Abel's inequality yields

   $|S_2|\le a_{N+1}\cdot\frac{2\pi}{x}=\frac{2\pi}{x\,(N+1)\ln (N+1)}.$

5. 5

   Finish the tail estimate. Because $N+1>1/x$, we have $x(N+1)>1$, so

   $|S_2|\le\frac{2\pi}{\ln(N+1)}\xrightarrow[x\to0^{+}]{}0.$

6. 6

   Combine. Both $S_1$ and $S_2$ tend to $0$ as $x\to0^{+}$, hence $f(x)\to0=f(0)$; oddness gives the same from the left. So $f$ is continuous at $x=0$. $\blacksquare$ The same estimates in fact prove more: since $n\,a_n=1/\ln n\to0$, the Chaundy–Jolliffe theorem says the sine series converges uniformly on all of $\mathbb{R}$, so $f$ is continuous everywhere.