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What do we require of $g(n)$, if for every positive strictly increasing unbounded $f(n)$, this sum converges?

$\sum_{n=1}^{\infty} \frac{\sin(g(n))}{f(n)} .$

Does it converge for $g(n)=n$?

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    I am 99% sure that you can not do better than Leibniz theorem - or a theorem equivalent to that. It is a bit messy to prove, but please look at my above comment and think what Leibniz would have done in the different cases.2011-10-04

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Let $S(n) = \sum_{k=0}^n g(k)$. As others have commented, if $S(n)$ is bounded, your sum converges. On the other hand, if $S(n)$ is unbounded, I will construct $f(n)$ such that your sum diverges. WLOG suppose $S(n)$ is unbounded above, and take an increasing function $N(m)$ on nonnegative integers so that $N(0) = 0$ and $S(N(m)) \ge m + S(N(m-1))$. Let $f_0(n) = m$ if $N(m-1) < n \le N(m)$. Then $\sum_{n=1}^{N(m)} \frac{\sin(g(n))}{f_0(n)} = \sum_{k=1}^m \ \sum_{n=N(k-1)+1}^{N(k)} \frac{\sin(g(n))}{k} = \sum_{k=1}^m \frac{S(N(k)) - S(N(k-1))}{k} \ge m$ This $f_0$ is not allowed only as $f$ because it is not strictly increasing. But a very slight adjustment will make it strictly increasing while still having, say, $\displaystyle\sum_{n=N(k-1)+1}^{N(k)} \frac{\sin(g(n))}{f(n)} \ge \frac{1}{2}$.