Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let lim $a_n=0$ and $s_N=\sum_{n=1}^{N}a_n$.

Show that $\sum a_n$ converges when $\lim_{N\to\infty}s_Ns_{N+1}=p$ for a given $p>0$.

I've no idea how to even start. Should I try to prove that $s_N$ is bounded ?

share|cite|improve this question
another typo, sorry guys ! – Kasper Feb 11 '13 at 15:46
Should I try to prove that $s_N$ is bounded... Good idea, did you do it? – Did Feb 11 '13 at 16:57
up vote 2 down vote accepted

Put $s_n:=\epsilon_n|s_n|$ with $\epsilon_n\in\{-1,1\}$. Then from $$\epsilon_n\epsilon_{n+1}|s_n|\>|s_{n+1}|=s_n\>s_{n+1}=:p_n\to p>0\qquad(n\to\infty)$$ it follows that $\epsilon_n=\epsilon_{n+1}$ for $n>n_0$. Assume $\epsilon_n=1$ for all $n> n_0$, the case $\epsilon_n=-1$ being similar.

The equation $$s_n(s_n+a_{n+1})=s_ns_{n+1}=p_n$$ implies that for all $n$ the quantities $s_n$, $a_{n+1}$, and $p_n$ are related by $$s_n={1\over2}\left(-a_{n+1}\pm\sqrt{a_{n+1}^2 +4p_n}\right)\ .$$ Since $s_n\geq0$ $\ (n>n_0)$, $\ a_{n+1}\to 0$, $\ p_n\to p>0$ it follows that necessarily $$s_n={1\over2}\left(-a_{n+1}+\sqrt{a_{n+1}^2 +4p_n}\right)\qquad(n>n_1)\ ,$$ and this implies $\lim_{n\to\infty} s_n=\sqrt{p}$.

share|cite|improve this answer
I don't understand this part: The equation $$s_n(s_n+a_{n+1})=s_ns_{n+1}=p_n$$ implies $$s_n={1\over2}\left(-a_{n+1}\pm\sqrt{a_{n+1}^2 +4p_n}\right)\ .$$ – Kasper Feb 13 '13 at 16:11
ooh you're using abc formula, but can you use this ? $a_{n+1}$ and $p_n$ are not constants right ? – Kasper Feb 13 '13 at 16:16
@Kasper: Of course; see my edit. – Christian Blatter Feb 13 '13 at 16:33
ooh okay, I see, thank you for this answer ! really aprreciated – Kasper Feb 13 '13 at 16:33

Your Answer


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.