# 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??

• Are $a_n, b_n$ real? If so, then hint: Cauchy-Schwarz. Feb 12, 2013 at 2:37
• I think you should double-check your second task. If $\sum a$ and $\sum b$ both converge, there is no proof that $\sum ab$ converges because it's not always true. If it were always true, then you would not be able to find the counter-example that you are asked to find for your third question. Are you sure the second question is not meant to be about two absolutely convergent series (as opposed to conditionally convergent)? Then the third part could be fulfilled by two conditionally convergent series. Feb 12, 2013 at 2:39
• I suspect the second part is: $\sum_{k=1}^\infty |a_n|, \sum_{k=1}^\infty |b_n|$ both converges, then $\sum_{k=1}^\infty a_n b_n$ converges. Feb 12, 2013 at 2:41
• To add to @ToddWilcox's comments, remember that it's possible for a sequence to go to zero, but for its sum to diverge, such as $a_k = 1/k$. Feb 12, 2013 at 2:42

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.

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.
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