How to prove if $a_n$ and $b_n$ converge and $a_n,b_n \ge 0$ then $a_n b_n$ also conveges?. If $a_n$ and $b_n$ are sequences of non-negative numbers and both converge, how do I argue that $\sum a_n b_n$ also converges?
My attempt:
I tried to argue that $a_n b_n\le(a_n+b_n)^2=a_n^2+2a_nb_n+b_n^2$, but I am stuck on how to proceed from here.. I have a feeling that I should use the comparison test, but I am not too sure.
 A: The condition stated in your post isn't clear. You could have $a_n = b_n = \dfrac{1}{\sqrt{n}}\to 0$, but $\sum_{n=1}^\infty a_nb_n = \sum_{n=1}^\infty \dfrac{1}{n} = \infty$. Thus the said conditions about $a_n$ and $b_n$ should be:  $a_n$ converges, $b_n$ converges, and either $\sum_{n=1}^\infty a_n$ or $\sum_{n=1}^\infty b_n$ converges. We can proceed from this point. Let's say $\sum_{n=1}^\infty a_n$ converges. Since $b_n$ converges,it is a bounded sequence, i.e. $|b_n| \le K, \forall n \ge 1$. We have: $0 \le a_nb_n \le Ka_n\implies \sum_{n=1}^\infty a_nb_n \le K\sum_{n=1}^\infty a_n$. This shows the series in question converges by comparison test.
A: 一）If an and bn are sequences of non-negative numbers and both converge, how do I argue that sum ∑anbn  also converges?
you can disprove it by plug in an=1/√n ,  bn=1/√n
∑aₙbₙ=∑1/n⇒+∞, is not a convergent sequence.
二)However, if the problem is:
If ∑an and ∑bn are sequences of non-negative numbers and both converge, how do I argue that sum ∑anbn  also converges?
proof:
for any N
LET ∑aₙbₙ=a៷b៷+...aₙbₙ
<（a៷+a៷₊₁+...aₙ)(b៷+...+bₙ)
because ∑an ,∑bn are convergent sequence,
thus for any ε>0, there exist  N,
∑an=a៷+a៷₊₁+...aₙ+...<ε
∑bn=b៷+...+bₙ+...<ε
hence ∑aₙbₙ=a៷b៷+...aₙbₙ+...<a៷b៷+...aₙbₙ<
（a៷+a៷₊₁+...aₙ)(b៷+...+bₙ)<(a៷+a៷₊₁+...aₙ+...)(b៷+...+bₙ+...)
<ε²=β
So.that.we have conclude that for any β>0, there is N
partial sum ∑aₙbₙ=a៷b៷+...aₙbₙ+...<β
so  ∑aₙbₙ is a Cauchy sequence, and it converge
