Does $\sum_{n=2}^{\infty}\frac{\sin n}{n \ln n}$ converge or diverge?
Answer
We use Dirichlet's test (https://en.wikipedia.org/wiki/Dirichlet%27s_test) to show that this infinite series converges. Since $\frac{1}{n \ln n}$ is decreasing for ($n\geq 2$) and bounded, it is enough to show that
\[S_n=\sum_{k=2}^{n}\sin k\]
is bounded. Notice that
\[S_n=\text{Real}(\sum_{k=2}^{n}e^{ik})=\text{Real}(\frac{e^{2i}e^{(n+1)i}}{1e^{i}}).\]
Hence
\[S_n \leq \frac{e^{2ie^{(n+1)i}}}{1e^{i}}=e^{2i}\frac{1e^{(n1)i}}{1e^{i}}\leq \frac{2}{1e^{i}}<\infty.\]
Thus it follows from the Dirichlet's test that
\[\sum_{n=2}^{\infty}\frac{\sin n}{n\ln n}\]
is convergenet.
Daniel90
436
The answer is accepted.
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