Determine if the following series is convergent:

$$\sum_{k=1}^\infty \frac{\sin^3 k}{\sqrt k}.$$

At first when I am solving the problem, I thought of the limit comparison test, if I divide the term by $\frac{k^3}{\sqrt k}$. Then I will get a limit of 1, and since the summation of the term I divide is divergent, I conclude that it is divergent.

However, I think that there is something wrong because if I do the same thing on $\frac{\sin^3 k}{k^2}$ by dividing it by $\frac{k^3}{k^2}$, I also get a limit of 1 but summation of k clearly diverges. This is not right because sum $\frac{\sin^3 k}{k^2}$ clearly converges. What is wrong here?

  • $\begingroup$ the series is divergent $\endgroup$ – tattwamasi amrutam Apr 26 '14 at 12:35
  • $\begingroup$ @TattwamasiAmrutam No it's not. $\endgroup$ – David H Apr 26 '14 at 12:39
  • $\begingroup$ It is convergent. $\endgroup$ – Claude Leibovici Apr 26 '14 at 12:59

The series is convergent: we may express $\sin^3j$ as a linear combination of $\sin(3j)$ and $\sin j$, so we are reduced to prove that the series $$\sum_{j=1}^\infty\frac{\sin(3j)}{\sqrt{j}}\mbox{ and }\sum_{j=1}^\infty\frac{\sin j}{\sqrt j}$$ are convergent.

To this aim, we use a summation by part plus boundedness of the sequences $\left(\sum_{j=1}^n\sin(3j)\right)_{n\geqslant 1}$ and $\left(\sum_{j=1}^n\sin j\right)_{n\geqslant 1}$.

  • $\begingroup$ what is wrong with the limit comparison test? $\endgroup$ – user10024395 Apr 26 '14 at 13:00
  • $\begingroup$ It is not true that $\sin^3k/k^3$ converges to $1$ as $k$ goes to infinity. $\endgroup$ – Davide Giraudo Apr 26 '14 at 13:19
  • $\begingroup$ i thought (sin k)/k tends to 1 so (sin k/k)^3 tends to 1? No? $\endgroup$ – user10024395 Apr 26 '14 at 13:24
  • $\begingroup$ It's true as $k$ goes to $0$, but not when $k$ goes to infinity. $\endgroup$ – Davide Giraudo Apr 26 '14 at 13:33
  • $\begingroup$ i think we just have to prove sin j / sqrt(j) and use comparison test, but I don't know how to prove it. $\endgroup$ – user10024395 Apr 27 '14 at 2:53

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