# Does $\sum_{n=2}^{\infty}\frac{\sin n}{n \ln n}$ converge or diverge?

## Answer

\[S_n=\sum_{k=2}^{n}\sin k\]

is bounded. Notice that

\[S_n=\text{Real}(\sum_{k=2}^{n}e^{ik})=\text{Real}(\frac{e^{2i}-e^{(n+1)i}}{1-e^{i}}).\]

Hence

\[S_n \leq |\frac{e^{2i-e^{(n+1)i}}}{1-e^{i}}|=|e^{2i}\frac{1-e^{(n-1)i}}{1-e^{i}}|\leq \frac{2}{|1-e^{i}|}<\infty.\]

Thus it follows from the Dirichlet's test that

\[\sum_{n=2}^{\infty}\frac{\sin n}{n\ln n}\]

is convergenet.

Daniel90

443

The answer is accepted.

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