Changing divergent series to convergent by re-ordering denominators Suppose $a_n$ is strictly decreasing and positive and $\sum_{n>1}a_n/n=\infty$, let $g:\mathbb N\to\mathbb N$ be a bijection between the positive integers, can we have $\sum_{n>1}a_n/g(n)<\infty$?
 A: If $a_n$ goes to zero, choose a subsequence whose $n$th term is smaller than $1/n$.
Now, for any index that is not a power of two, pair up $1/n$ with the term of the subsequence that is smaller than $1/n$.
The sum of these terms is smaller than $\sum \frac 1 {n^2}=\pi^2/6$.
Pair up the remaining $a_n$ with the remaining $1/2^k$. This gives a series that is smaller than $\sum a_1 \frac 1 {2^n}=a_1$. (The $a_n$ are decreasing and bounded by $a_1$.)
Since everything is positive, this implies that the series converges.
A: We may assume, $a_n\searrow 0$ (otherwise, you can't do it).
Choose a subsequence $\{a_{n_k}\}$ with $a_{n_k}\le {1\over 2^k}$.
We write our new sequence: $a_1/2, a_2/4, \ldots, a_{n_1-1}/2^{n_1-1}$
The next term is $a_{n_1}/1$.
For terms after $a_{n_1}$ and before $a_{n_2}$ we continue dividing by powers of 2.
The next term is $a_{n_2}/ 3$.
In general: 
Terms $a_i$ that are not a term of the subsequence are divided by   $2^i$. The series formed by these terms will converge since the $a_i$ are decreasing and $\sum{1\over 2^n}$ converges
A term $a_{n_k}$ is divided by the first integer that hasn't been used to that point.
The series form by these terms clearly converges, since $a_{n_k}\le 2^{-k}$ and we made the terms smaller.
Thus, the resulting series of nonnegative terms converges.
I think this works. I'd try to formalize it; but I'm sure I would just make a mess of things...
