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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 ?

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another typo, sorry guys ! –  Kasper Feb 11 '13 at 15:46
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Should I try to prove that $s_N$ is bounded... Good idea, did you do it? –  Did Feb 11 '13 at 16:57

1 Answer 1

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 $\lim_{n\to\infty} s_n=\sqrt{p}$.

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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

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