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The series is $$\sum_{n=1}^\infty\frac{\sin n\cdot\sin n^2}{n}$$ It seems to use Dirichlet's test, but I cannot prove $\sum\sin n\cdot\sin n^2$ is bounded. This question may help -- Convergence of $\sum \limits_{n=1}^{\infty}\sin(n^k)/n$

user21820
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nsigma
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1 Answers1

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Approach 1: Telescoping Sum $$ \begin{align} \sum_{k=1}^n\sin(k)\sin\left(k^2\right) &=\frac12\sum_{k=1}^n(\cos(k(k-1))-\cos(k(k+1)))\\ &=\frac{1-\cos(n(n+1))}2\tag1 \end{align} $$ Thus, the partial sums are bounded by $1$. Dirichlet's Test and $(1)$ say that the original series converges.

Approach 2: Summation by Parts $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(k)\sin\left(k^2\right)}k &=\lim_{n\to\infty}\sum_{k=1}^n\frac{\sin(k)\sin\left(k^2\right)}k\\ &=\lim_{n\to\infty}\frac12\sum_{k=1}^n\frac{\cos(k(k-1))-\cos(k(k+1))}k\\ &=\lim_{n\to\infty}\left(\frac12-\frac{\cos(n(n+1))}{2n}-\frac12\sum_{k=1}^{n-1}\frac{\cos(k(k+1))}{k(k+1)}\right)\\ &=\frac12-\frac12\sum_{k=1}^\infty\frac{\cos(k(k+1))}{k(k+1)}\tag2 \end{align} $$ and the last sum converges by comparison to $$ \sum_{k=1}^\infty\frac1{k(k+1)}=1\tag3 $$

robjohn
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    Why does this imply the original series converge? – NonalcoholicBeer Jun 22 '18 at 13:21
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    @ablmf: Since the sum in this answer is bounded, Dirichlet's Test says the sum in the question converges. – robjohn Jun 22 '18 at 13:29
  • Impressive, I could have bet the sum diverges. It doesn't look that different from the harmonic series. – Eric Duminil Jun 22 '18 at 16:05
  • @EricDuminil: the terms oscillate, sort of like the alternating harmonic series (which converges to $\log(2)$). – robjohn Jun 22 '18 at 16:09
  • @robjohn: D'oh! I didn't think about that. It makes perfect sense now! – Eric Duminil Jun 22 '18 at 16:23
  • I really think, @robjohn, that you cash in on your status as a moderator to garner rep for answering some of the poorest questions asked on this site. Can you even answer a question that isn't a PSQ and/or homework? I can't even remember, because you answer so many basic homework questions and "Here's the question directly copied, please do my work for me" questions. – amWhy Jun 23 '18 at 01:49
  • If you feel that my above assessment is wrong, then show me otherwise, please! – amWhy Jun 23 '18 at 01:50