Use summation by parts to show that \(\sin(n)/n\) converges.
Solution
As written, \(\sin(n)/n\) is a sequence rather than a series, and its convergence is immediate: \[ \left|{\sin n\over n}\right|\le {1\over n}\longrightarrow0. \] Thus the displayed sequence converges to \(0\); summation by parts is not needed for this assertion.
The standard summation-by-parts exercise suggested by the wording is the convergence of \[ \sum_{n=1}^\infty {\sin n\over n}. \] For that corrected statement, put \[ B_N=\sum_{n=1}^N\sin n. \] Using a geometric sum, \[ \sum_{n=1}^N e^{in} =e^i{1-e^{iN}\over1-e^i}, \] so \((B_N)\) is bounded. Summation by parts gives, for \(N>M\), \[ \sum_{n=M}^N {\sin n\over n} ={B_N\over N}-{B_{M-1}\over M} +\sum_{n=M}^{N-1}B_n\left({1\over n}-{1\over n+1}\right). \] If \(|B_n|\le C\), the absolute value of the right-hand side is at most \[ {C\over N}+{C\over M} +C\sum_{n=M}^{N-1}\left({1\over n}-{1\over n+1}\right) \le {3C\over M}, \] which tends to \(0\) as \(M\to\infty\). Hence the corrected series is Cauchy and therefore convergent.