Take the 2-minute tour ×  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 $\sum_{n>1}a_n/g(n)<\infty$?  share|improve this
question  2 Answers 2  up vote 6 down vote accepted  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.  share|improve this answer The idea is to form
one subsequence that uses "few" $a_n$ and "many" $1/n$ so that
$a_n$ is small and makes the product small, and another
subsequence that uses "many" $a_n$ an