Let {$a_n$} and {$b_n$} be sequences such that $\sum_{n=1}^\infty a_n^2$ and $\sum_{n=1}^\infty b_n^2$ are convergent
Prove that $\sum_{n=1}^\infty a_nb_n \le (\sum_{n=1}^\infty a_n^2)^{\frac{1}{2}} (\sum_{n=1}^\infty b_n^2)^{\frac{1}{2}}$
I can see I have to use the Cauchy-Schwarz inequality. But not really sure how to start. Also I know $\sum_{n=1}^\infty a_nb_n$ converges absolutely from a previous part of the question.
Best Answer
Hint: Let $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$. What does Cauchy-Schwarz say? The dot product of $a\cdot b \leq (a\cdot a)^{1/2}(b\cdot b)^{1/2}$. Note, that the terms in the brackets correspond to your expressions. Then let $n\to \infty$.