# Proving the existence of a non-decreasing sequence

If $\{a_{n}\}$ is a sequence of positive numbers so that $\sum_{n=0}^{\infty} a_{n}$ converges. How can we show that there exists a non-decreasing sequence $\{b_{n}\}$ so that $\lim_{n\to \infty}b_{n}=\infty$ and $\sum_{n}^{\infty}a_{n}b_{n}$<$\infty$.

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$\{b_n\}$ modifies the terms of the sequence. The trick is to modify chunks of the sum into terms of another convergent sequence. $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent, so we aim for this.

As $\sum_{i=0}^{\infty}a_i$ is convergent, there is a $k_n$ for all $n$ such that $\sum_{i=k_n}^{\infty}a_i<\frac{1}{(n)(n^2)}$. Then set $b_i=n$ for $i=k_n$ to $i=k_{n+1}-1$ for $n\geq1$ and $b_i=0$ otherwise.

Then $\sum_{i=k_n}^{k_{n+1}-1}a_ib_i<\frac{1}{n^2}$ for $n \geq 1$. Summing all these chunks gives a convergent sequence. And $b_n$ tends to $\infty$ as required.

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$${a_n} = \frac{1}{ a^{3n }}$$

$${b_n} = a^{n}$$

for $a>1$, I think your requirements meet.

more explanation : in order to ensure $\sum_{n}^{\infty} a_n$to converge,

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} < p$$ for some $0<p<1$

find $b$ that

$$1< \lim_{n\to\infty} \frac{b_{n+1}}{b_n} < \frac{1}{ p}$$

so$\sum_{n}^{\infty} b_n$ would diverge while nondecreasing, and

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}\frac{b_{n+1}}{b_n} < \frac{p}{ p} = 1$$

which means $\sum_{n}^{\infty} a_n b_n$ would converge

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Yes. There exists a sequence for which the sum diverges. But that's not a counterexample to the question. – Asaf Karagila Nov 19 '12 at 10:31
if $a=2$, I think it is the solution you're looking for.. – thkang Nov 19 '12 at 10:32
Your answer gives an example of $a_n$ and gives an example of $b_n$. This is not what the question asks for. The questions says $a_n$ is a given sequence. Now find one example of $b_n$ such that $\lim b_n=\infty$ and $\sum a_nb_n<\infty$. – Asaf Karagila Nov 19 '12 at 10:35
What do you mean by one example of ${b_n}$ ? judging from the question, isn't my trivial $a_n$ a sequence of positive number whose sum converges? – thkang Nov 19 '12 at 10:41
@thkang: The thing what Asaf is trying to point out, is that $(a_{n})$ is an arbitrary sequence satisfying the required conditions. You can't prove this theorem by choosing some particular sequence $(a_{n})$ and finding $(b_{n})$ respectively. By doing this, it only proves that it works in this special case. – T. Eskin Nov 19 '12 at 10:49

If $\sum_{n}^{\infty} a_n$ converges then (I think) you can show that $\limsup|\frac{a_{n+1}}{a_n}| = p$ where $0 < p <= 1$. If this is true then you need to find a sequence $b_n$ which tends really slowly to infinity - so slowly in fact that $\sum_{n}^{\infty} a_nb_n$ converges. Finding a specific $b_n$ in terms of $p$ shouldn't be that hard.

For example, in the case where 0 < p < 1, you can just let b_n = (1-p)(a_n).

Asaf has shown that the case where p = 1 is trickier, so let's see if we can make progress... even if we can't it's still instructive to think about this method

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No, $p=1$ is also possible. For example $a_n=\frac1{n^2}$. – Asaf Karagila Nov 19 '12 at 10:50
Asaf - In that case, finding (b_n) should be even easier because a_n tends to 0 even faster. – Adam Rubinson Nov 19 '12 at 10:57
But you can replace $2$ by $1+\varepsilon$ for every $\varepsilon>0$, which makes it possible to find arbitrarily slow sequences. – Asaf Karagila Nov 19 '12 at 10:58
If epsilon is fixed between 0 and 1 then my method still works. If you're talking about something like: let epsilon tend slowly to 0 as n tends to infinity, then in that case I'm not sure Sum(a_n) will converge any more. Unless you can provide me with some example (which will probably involve logs) – Adam Rubinson Nov 19 '12 at 11:10

Fixed some $\alpha\in(0,1)$, let $\displaystyle R_n=\sum_{k=n}^\infty a_k,\;\; b_n=\frac{1}{R_n^\alpha}$. It is easy to see that $b_n$ is non-decreasing with $\displaystyle \lim_{n\to\infty}b_n=\infty$. We have

$\displaystyle a_nb_n=\frac{R_n-R_{n+1}}{R_n^\alpha}<\int_{R_{n+1}}^{R_n}\frac{1}{x^\alpha}dx=\frac{R_{n}^{1-\alpha}-R_{n+1}^{1-\alpha}}{1-\alpha}$

Hence, $\displaystyle \sum_{n=1}^\infty a_nb_n <\frac{R_1^{1-\alpha}}{1-\alpha}=\frac{S^{1-\alpha}}{1-\alpha}<\infty$

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b_n does not tend to infinity – Adam Rubinson Nov 19 '12 at 12:24
@Adam Rubinson Of course it does. Since $\sum\limits_{n=1}^\infty a_n$ converges, which implies that $\lim\limits_{n\to\infty}R_n = 0$. – Eastsun Nov 19 '12 at 13:02
but R_n does not tend to 0. R_n is a positive real number. – Adam Rubinson Nov 19 '12 at 13:32