Show if $\sum\limits_{k=1}^\infty {a_k}^2$,$\sum\limits_{k=1}^\infty {b_k}^2$ converge, their product converges too Show if $\displaystyle\sum\limits_{k=1}^\infty {a_k}^2$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}^2$ both converge, $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ also converge
then
Show if $\displaystyle\sum\limits_{k=1}^\infty {a_n}$ and $\displaystyle\sum\limits_{k=1}^\infty {b_n}$ both converge, $\displaystyle\sum\limits_{k=1}^\infty {a_nb_n}$ also converge
then
Find two convergent series $\displaystyle\sum\limits_{k=1}^\infty {a_k}$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}$ such that $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ diverges.

My thought and attempt:
For the first part, By the theorem (If the series $\displaystyle\sum\limits_{k=1}^\infty {a_n}$ is convergent, then $\displaystyle\lim_{n\to\infty} a_n=0$),
if $\displaystyle\sum\limits_{k=1}^\infty {a_k}^2$ and $\displaystyle\sum\limits_{k=1}^\infty {b_k}^2$ both converge, then $\displaystyle\lim_{n\to\infty} {a_k}^2=0$ and $\displaystyle\lim_{n\to\infty} {b_k}^2=0$ both equal to zero.
Then we can obtain both $(a_k)_{k=1}^\infty$ and $(b_k)_{k=1}^\infty$ converge
thus $\displaystyle\lim_{n\to\infty} {(a_k)^2\over|a_k|} = \displaystyle\lim_{n\to\infty} |a_k| = 0$, do the same thing for $b_k$
Since limit preserves arithmetic operation, $\displaystyle\sum\limits_{k=1}^\infty {a_kb_k}$ converges.
Second part, I dont have any idea...
Third part is $a_n = b_n = {(-1)^n\over\sqrt{n+1}}$, am I right??
Thank you everyone... please help...
 A: Since $(a-b)^2 \geq 0$, you have $|ab| \leq \frac{1}{2}(a^2+b^2)$.
Hence you have $\sum_{k=1}^n |a_k b_k| \leq \frac{1}{2} \sum_{k=1}^n (a_k^2+b_k^2) \leq \frac{1}{2} \sum_{k=1}^\infty (a_k^2+b_k^2)$, from which it follows that $\sum_{k=1}^\infty |a_k b_k| < \infty$.
If $\sum_{k=1}^\infty |a_k| < \infty$  then $a_k \to 0$. In particular, the terms are bounded, ie, $|a_k| \leq M$ for some $M$. Then since $a_k^2 \leq M |a_k|$, it follows that $\sum_{k=1}^\infty a_k^2 < \infty$. Similarly $\sum_{k=1}^\infty b_k^2 < \infty$, and using the first result yields the desired result.
However, if the sequences are not absolutely convergent, the result is not true. Take $a_k = b_k = (-1)^k \frac{1}{\sqrt{k}}$. These series converge because they are alternating and the terms converge to zero, but $a_k b_k = \frac{1}{n}$, and it is well known that $\sum_n \frac{1}{n}$ is divergent.
A: By the Cauchy-Schwarz Inequality,
$$
\forall n \in \mathbb{N}: \quad \sum_{k=1}^{n} |a_{k} b_{k}| \leq \sqrt{\sum_{k=1}^{n} a_{k}^{2}} \cdot \sqrt{\sum_{k=1}^{n} b_{k}^{2}}.
$$
Therefore,
$$
\sum_{k=1}^{\infty} |a_{k} b_{k}| \leq \sqrt{\sum_{k=1}^{\infty} a_{k}^{2}} \cdot \sqrt{\sum_{k=1}^{\infty} b_{k}^{2}} < \infty.
$$
In the language of real analysis, what we have shown is the following.

Theorem If $ (a_{k})_{k \in \mathbb{N}},(b_{k})_{k \in \mathbb{N}} \in {\ell^{2}}(\mathbb{N}) $, then $ (a_{k} b_{k})_{k \in \mathbb{N}} \in {\ell^{1}}(\mathbb{N}) $ and
  $$
\left\| (a_{k} b_{k})_{k \in \mathbb{N}} \right\|_{1} \leq \left\| (a_{k})_{k \in \mathbb{N}} \right\|_{2} \cdot \left\| (b_{k})_{k \in \mathbb{N}} \right\|_{2} < \infty.
$$


Just to add to copper.hat’s last example.

Both $ \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} $ and $ \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k}}{\ln(k)} $ converge by the Alternating Series Test, but $ \displaystyle \sum_{k=1}^{\infty} \frac{1}{k \ln(k)} $ diverges by the Integral Test.

