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 $\sum_{n}^{\infty}a_{n}b_{n}$<$\infty$.
 A: $\{b_n\}$ 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.
A: 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
A: 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, $\displaystyle \sum_{n=1}^\infty a_nb_n <\frac{R_1^{1-\alpha}}{1-\alpha}=\frac{S^{1-\alpha}}{1-\alpha}<\infty$
A: $$ {a_n} = \frac{1}{ a^{3n }} $$
$${b_n} = a^{n} $$
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
