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If $f(x)>0$ is any function st. $\sum_{j=1}^{\infty}f(j)=\infty$. And $a_n=f(n)$ with probability 50%, else 0.

Does $\lim_{n\rightarrow\infty}((1/\sum_{j=1}^nf(j)) \sum_{k=1}^n a_k)=1/2$ almost surely?

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Did they show your pageant on television? I missed it. – Will Jagy Jan 23 '12 at 0:36
I see, after you revised your user name: the pageant is later in the year and you are very confident. Good luck. – Will Jagy Jan 23 '12 at 1:11
@WillJagy May I know "What's in a name? That which we call a rose By any other name would smell as sweet." – user21436 Jan 23 '12 at 1:21
@KannappanSampath There was an Irish author/playwright who was wonderful at inventing names, he is usually just referred to as Lord Dunsany,,_18th_Baron_of_Dunsany Here is the entire text of his first book: – Will Jagy Jan 23 '12 at 1:31
up vote 2 down vote accepted

No, not for every $f$. If you choose $f$ growing so rapidly that $f(n)>3\sum_{j=1}^{n-1}f(j)$, then every time that $a_n=f(n)$ we get $(\sum_{k=1}^n a_k)/\sum_{j=1}^nf(j)\geq f(n)/\sum_{j=1}^nf(j) >3/4$. And this happens infinitely often with probability 1.

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Call $F_n=\sum\limits_{k=1}^nf(k)$ and assume that $\frac{F_{n+1}}{F_n}\to c$ for a given $c\geqslant1$ and that $\frac1{F_n}\sum\limits_{k=1}^na_k$ converges almost surely to a limit $\ell$. Then, $\frac1{F_n}\sum\limits_{k=1}^{n+1}a_k=\frac{F_{n+1}}{F_n}\cdot\frac1{F_{n+1}}\sum\limits_{k=1}^{n+1}a_k$ converges almost surely to $c\ell$, hence $\frac{a_{n+1}}{F_n}=\frac1{F_n}\sum\limits_{k=1}^{n+1}a_k-\frac1{F_n}\sum\limits_{k=1}^{n}a_k$ converges almost surely to $(c-1)\ell$. Since $a_{n+1}=0$ infinitely often almost surely, this shows that $(c-1)\ell=0$.

Hence, $\frac1{F_n}\sum\limits_{k=1}^na_k$ cannot converge almost surely to a nonzero limit unless $\frac{F_{n+1}}{F_n}\to1$.

If $f(n)=c^n$ with $c\gt1$, then $\frac{F_{n+1}}{F_n}\to c\ne1$ hence $\frac1{F_n}\sum\limits_{k=1}^na_k$ does not converge almost surely to any positive limit.

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