Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g(n)=n$?

share|improve this question
    
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
4  
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
add comment

1 Answer 1

up vote 3 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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