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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 $\sum_{n>1}a_n/g(n)<\infty$?

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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.

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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" $a_n$ and "few" $1/n$ so that $1/n$ is small and makes the product small. This is the same idea as in David's answer, of course. –  Phira Nov 23 '11 at 19:10
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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...

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hum, but isn't the whole problem that the terms that are not terms of the subsequence may be very, very numerous ? it doesn't matter that you divide them by powers of 2 (say less than 2^n_k) if there are way more than 2^n_k of them... For this reason I actually think it is false (but definitely not trivial !) –  Glougloubarbaki Nov 23 '11 at 19:09
Those terms are bounded, and the series formed by those terms would converge since $\sum 2^{-n}$ converges. –  David Mitra Nov 23 '11 at 19:14
it's hard to say it precisely, but if you take $a_n = 1/(\ln(1+ \ln(1+ \ln(1+ n))))$ the number of terms before $a_n < 1/4$ is monstruous, and you'll end up with a sum of an enourmous number of terms of the size $1/4 /(\ln(1+ \ln(1+ \ln(1+ n))))$ which is not so small, and so on at every stage, so that the sum will not converge. –  Glougloubarbaki Nov 23 '11 at 19:15
I am also led to a similar conclusion. If it needs to formalize, one may do like this: Suppose $a_n \downarrow 0$. Let $(x_{n})$ be an increasing sequence so that both $a_{x(k)} \leq 2^{-k}$ and $x(k) \geq 2^{k}$. Also let $(y_{n})$ be the enumeration of the set $\mathbb{N} - \{ x_n : n \in \mathbb{N} \}$ in increasing order. Then define $$g(n) = \begin{cases} x(k) & \text{if} \ n = y(k) \\ y(k) & \text{if} \ n = x(k) \end{cases}.$$ Then $$\sum_{n \geq 1} \frac{a_n}{g(n)} = \sum_{k \geq 1} \frac{a_{x(k)}}{y(k)} + \sum_{k \geq 1} \frac{a_{y(k)}}{x(k)},$$ which clearly converges. –  sos440 Nov 23 '11 at 19:18
ok, you're right –  Glougloubarbaki Nov 23 '11 at 19:25
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