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 $\sum_{n}^{\infty}a_{n}b_{n}$&lt;$\infty$.
$\{b_n\}$ 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 $\sum_{i=k_n}^{\infty}a_i&lt;\frac{1}{(n)(n^2)}$. Then set $b_i=n$ for $i=k_n$ to $i=k_{n+1}-1$ for $n\geq1$ and $b_i=0$ otherwise.

Then $\sum_{i=k_n}^{k_{n+1}-1}a_ib_i&lt;\frac{1}{n^2}$ for $n \geq 1$. Summing all these chunks gives a convergent sequence. And $b_n$ tends to $\infty$ as required.
$$ {a_n} = \frac{1}{ a^{3n }} $$

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

for $ a&gt;1 $, I think your requirements meet.

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

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

find $ b$ that 

$$ 1&lt; \lim_{n\to\infty} \frac{b_{n+1}}{b_n} &lt; \frac{1}{ p}  $$

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

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

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