3
$\begingroup$

I have the following series: $$ \sum\limits_{n=1}^{\infty} \frac{\sin(n)}{3n^p-\sin(n)} $$ and I want to show that this series diverges for $0<p \leqslant \frac 1 2$. I tried to write $$ \frac{\sin(n)}{3n^p - \sin(n)} = \frac{\sin(n)}{3n^p} \left(1-\frac{\sin(n)}{3n^p}\right)^{-1} = \frac{\sin(n)}{3n^p}\left(1+\frac{\sin(n)}{3n^p}+\cdots \right) \\ = \frac{\sin(n)}{3n^p} \left(1+\frac{\sin(n)}{3n^p}+\cdots+\frac{\sin^k(n)}{(3n^p)^{k}} \left(1-\frac{\sin(n)}{3n^p}\right)^{-1}\right) \\ = \frac{\sin(n)}{3n^p} + \frac{\sin^2(n)}{(3n^p)^{2}}+\cdots+\frac{\sin^k(n)}{(3n^p)^k} + \frac{\sin^{k+1}(n)}{(3n^p)^k \bigl( 3n^p-\sin(n) \bigr)} $$ and then choose for any fixed $p$ such $k$ that $pk > 1$ so that the series related to last term converge. The series related to first term converges for any $p>0$. So there is a problem to show that the series, related to group of terms from first to $k$-th diverges. For each of these terms I can show this by expanding $\sin^m(n)$ as a sum of sines and cosines in first power and constants. But how can I show that the resulting sum will correspond to an also divergent series? Maybe there is some better way to show that my series diverges?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ The terms with odd $k$ converge (by Abel-Dirichlet), the terms with even $k$ are positive, so it's enough to show that the $k=2$-part diverges. For that one needs to show that even if you remove $n$'s with $\sin^2 n<\epsilon$, the harmonic series is divergent. It seems 'obvious', but I needs a proof. $\endgroup$ – user8268 Aug 30 '13 at 19:58
3
$\begingroup$

Summation by parts gives the convergence of $\sum_n\frac{\sin n}{n^p}$, so we have to deal with the convergence of $\sum_n a_n$, where $a_n:=\sin n\left(\frac 1{3n^p-\sin n}-\frac 1{3n^p}\right)=\frac{\sin^2n}{3n^p(3n^p-\sin n)}$.

We have a series with non negative terms, and $a_n\sim\frac{\sin^2n}{9n^{2p}}$.

Rewriting $\sin^2n$ as $1-\cos(2n)$ (up to a multiplicative constant), noticing that $\sum_n\frac{\cos(2n)}{n}$ is convergent and $\sum_n\frac 1{n^{2p}}$ is divergent, we conclude the divergence of the initial series.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.