I have a question I thought I had the answer to, but I made some false assumptions. Say we have $a_{n}=\sum_{k=1}^{n} c_{n}$ and we know that $\frac{c_{k+1}}{c_k}$ is unbounded, does $a_{n}$ converge or diverge? We know that $\forall M \in \mathbb{R},\exists N:\frac{c_{N+1}}{c_N}>M$. I imagine the sequence to diverge but how to prove this, I don't know. Since if $a_{n}$ would converge it would mean that $c_{n} \rightarrow 0$ but that would mean that $c_{n+1}$ would have to become infinitely large in order to have $\frac{c_{N+1}}{c_N}>M$ still valid... I'm missing something, can anybody help?


Consider $c_{2n-1} = \dfrac{1}{2^n}, c_{2n} = \dfrac{1}{n^2}$. $\displaystyle\sum_{n=1}^\infty c_n$ is absolutely convergent to $1 + \dfrac{\pi^2}{6}$. However, $\dfrac{c_{2n}}{c_{2n-1}} = \dfrac{2^n}{n^2}$, which is unbounded.

  • $\begingroup$ I see what you did, but I already proved this (although slightly different). But this isn't stating anything about the ratio being unbounded, and that is what I'm looking for. In other words: if I can prove that if $\frac{c_{n+1}}{c_{n}}$ being unbounded means that $c_{n}$ will not converge to 0, then I'll have a solid proof. Do you see what I mean? $\endgroup$ – Mathbeginner Dec 4 '17 at 21:36
  • $\begingroup$ If $\dfrac{c_{n+1}}{c_n}$ is unbounded, then so is $\left|\dfrac{c_{n+1}}{c_n}\right|$, yes? So $\displaystyle\lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right|=\infty > 1$ $\endgroup$ – Araske Dec 4 '17 at 21:41
  • $\begingroup$ The limit of an unbounded set doesn't have to go to infinity. Take for example {1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9.....}, this sequence is unbouded but doesn't go to infinity. I believe that holds for all oscillating sequences as well.. $\endgroup$ – Mathbeginner Dec 4 '17 at 21:43
  • $\begingroup$ Oh! Within the exact confines of your question, I think the series might well be allowed to converge. I apologize, I misread. $\endgroup$ – Araske Dec 4 '17 at 21:47
  • $\begingroup$ I've changed my answer to provide an example of a convergent such series, does this fit in the confines of your question? $\endgroup$ – Araske Dec 4 '17 at 21:55

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