If $\sum_{n=1}^\infty a_n$ converges, then there is a sequence with $b_n \to \infty$ and $\sum_{n=1}^{\infty} a_n b_n < \infty.$ Let $\{a_n\}_{n=0} ^\infty$ be a sequence of positive numbers so that $\sum_{n=1}^{\infty} a_n$ converges.
a) show that there exists an nondecreasing sequence $\{b_n\}$ so that $\lim_{n\to\infty}b_n = \infty$ and $\sum_{n=1}^{\infty} a_n b_n < \infty.$
How can I use the partial sums of $\sum_{n=1}^{\infty} a_n  $ to define $\{b_n\}$?
 A: Here is an idea? You can start with a constant $b_n=c_1$ for $n=1,2,\ldots$, but not forever, of course. Keep an eye on $\sum_{n=N}^\infty a_nc_1$. It goes to zero as $N\to\infty$. When the sum has become comfortably small, start with $b_n=c_2=2c_1$ for $n=N_1,N_1+1,\cdots$. Again, keep going until the tail of the new sum is really small, then double the $b_n$ again, putting $b_n=c_3=2c_2$ for $n=N_2,N_2+1,\cdots$. The point is using each constant for longer and longer, ensuring that the partial sums you are collecting along the way grow more and more slowly.
This is just a half-baked idea, but it should work.
A: Yet another idea. Put $$t_n=\sum_{k=n}^\infty a_k$$ and note that $t_n\to0$. But then $$\sum_{n=1}^\infty \frac{a_n}{t_n^p}<\infty\qquad\text{if $0<p<1$.}$$
To see this, write $a_n=t_n-t_{n+1}$ and compare the $n$th term in the sum with $$\int_{t_{n+1}}^{t_n}\frac{dt}{t^p}.$$
A: First condition is product(a(n)*b(n))-->0 as n-->infinity
Since a(n) and b(n) are functions of 'n' represent the product as f(n)
Sum [a(n)*b(n)] (from 1 to inf.) can be approximated by S=f(1)/2+integral[f(x)*dx]
(from 1 to inf.) If it can be shown ,that this Integral converges to a finite value ,then the Infinite sum also converges.
