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If $\pi=\sum_{n>0} a_n/n$, where $a_i=1$ iff $\pi>\sum_{1+i>n>0} a_n/n$, else 0.

What is $\sum_{n>0} a_n/n^2$?

Does $\sum_{n>0} a_n/\ln(n)$ and $\sum_{n>0} a_n/\sqrt n$ converge?

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    Motivation? ${ }$2012-02-24

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Let $S$ be the sum of the first few terms. Suppose $S+(1/n)\lt\pi\lt S+(1/(n-1))$, so $a_n=1$. Then $\pi-(S+(1/n))\lt1/(n(n-1))$, so the next $m$ with $a_m=1$ satisfies $m\gt n(n-1)$. Thus in the sequence of $n$ such that $a_n=1$, each term is bigger than (something almost as big as) the square of the previous $n$. So the logarithm of those $n$ is growing exponentially, and $\sum(a_n/\log n)$ converges. A fortiori, $\sum(a_n/\sqrt n)$ converges.

I doubt that any of these series converges to anything interesting. You could try calculating a few digits and then looking it up somewhere.