# Convergence of $\frac{\sin(g(n))}{f(n)}$

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'm not sure I understand the last question. Are you asking if the sum converges if $g(n) = n$ and $f(n)$ is an arbitrary strictly increasing unbounded function? – JavaMan Oct 4 '11 at 20:03
If $\sum_{k = 0}^n \sin(g(k))$ is bounded, then I think summation by parts shows your series converges. This works in the case of $g(n) = n$, because we have $$\sum_{k = 0}^n \sin(k) = \Im \sum_{k = 0}^n e^{ik} = \Im \frac{e^{i(n+1)}-1}{e^i-1}$$ And the latter is bounded. – Joel Cohen Oct 4 '11 at 20:05
Try to evaluate the sum with $f(n) = \log (n)$ and $g(n) = n$ (you will need to start with $n = 2$). – JavaMan Oct 4 '11 at 20:09
I would formulate this using $\sum a_nb_n$ with $b_n$ bounded and $a_n$ strictly decreasing to $0$. – AD. Oct 4 '11 at 20:16
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. – AD. Oct 4 '11 at 21:05

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, $\displaystyle\sum_{n=N(k-1)+1}^{N(k)} \frac{\sin(g(n))}{f(n)} \ge \frac{1}{2}$.