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I saw a question on quora asking whether or not the sum ${\sum}_{k=0}^{\infty}{sin(2^k)\over n}$ is convergent. My opinion, and that of the other answers, is that Dirichlet's test could be used with {$a_{n}$}=$1\over n$ and {$b_{n}$}=$\sin(2^n)$. I'm lost, however, when it comes to showing whether or not ${\sum}_{k=0}^{\infty}b_n\leq M$ for some $M$. It was mentioned that an integral test could be used, however, an integral test only works for monotonically decreasing functions, which $\sin(2^x)$ is clearly not.

For those who do not know Dirichlet's convergence test: https://en.wikipedia.org/wiki/Dirichlet%27s_test

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  • $\begingroup$ If a series converges, its partial sums form a Cauchy sequence. Do these partial sums form a Cauchy sequence? $\endgroup$
    – Dzoooks
    Commented Jun 14, 2018 at 2:10
  • $\begingroup$ What you mean is that the partial sums $\sum_{k=0}^n b_k$ are bounded. This is almost certainly not true. $\endgroup$
    – Robert Israel
    Commented Jun 14, 2018 at 2:12
  • $\begingroup$ Is this supposed to be the sum $$ \sum_{k=1}^\infty \frac{\sin 2^k}{k}$$ ? $\endgroup$
    – N8tron
    Commented Jun 14, 2018 at 3:02
  • $\begingroup$ Yes, ultimately the goal is to check the convergence of your mentioned sum (I have edited the question title to reflect this). However, trying to do this myself led me to question whether or not $\sum_{k=0}^{\infty}\sin(2^k)$ is bounded. An answer or hint to either of both of these questions would be much appreciated. $\endgroup$
    – R.Jackson
    Commented Jun 14, 2018 at 3:06
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    $\begingroup$ isn't this one of the convergence questions where you need crazy s*it like the irrationality measure of $\pi$ to come even close to a solution? $\endgroup$
    – tired
    Commented Jun 14, 2018 at 6:29

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