Let $a_n$ and $b_n$ be two sequences of real numbers such that series $\sum a_n^2$ and $\sum b_n^2$ converge. Then the series $\sum a_nb_n$

A. is absolutely convergent

B. may not converge

C.is always convergent, but maynot converge absolutely

D.converges to 0


I took $a_n =(-1)^n$ and $b_n=1$. Then , $a_nb_n$ is not convergent.So this removes options C and D.If i take $a_n=\frac{(-1)^n}{n}$ and $b_n=1$, it removes option A. So B is left. But i am not sure



marked as duplicate by Empty, user228113, Mark Bennet, Claude Leibovici, 3SAT Jan 3 '16 at 10:00

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    $\begingroup$ I think the problem is talking about convergence of series $\sum a_n^2$ and others, not sequences $a_n^2$ themselves. $\endgroup$ – Wojowu Jan 1 '16 at 13:40
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    $\begingroup$ Your examples are invalid: in both cases, $\sum b_n^2$ fails to converge. $\endgroup$ – Omnomnomnom Jan 1 '16 at 14:00
  • $\begingroup$ why does $\sum_k 1$ converge? $\endgroup$ – Alex Jan 1 '16 at 14:35

For all $a,b \in \mathbb{R}$ we have $$ |ab| = |a| |b| \leq 2 |a| |b| \leq |a|^2 + |b|^2 = a^2 + b^2. $$ Because $\sum_{n=0}^\infty a_n^2 < \infty$ und $\sum_{n=0}^\infty b_n^2 < \infty$ it follows that $$ \sum_{n=0}^\infty |a_n b_n| \leq \sum_{n=0}^\infty (a_n^2 + b_n^2) = \sum_{n=0}^\infty a_n^2 + \sum_{n=0}^\infty b_n^2 < \infty, $$ so the series converges absolutely.

PS: This shows that A. holds, and thus also answers if B. and C. hold. Notice that D. does not hold; take for example $a_0 = b_0 = 1$ and $a_n = b_n = 0$ for $n > 0$.

  • $\begingroup$ Note that you reordered the sum $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ to $\sum_{n=0}^\infty a_n^2 + \sum_{n=0}^\infty b_n^2$ which is only possible because the latter two series converge absolutely (because of the square) by assumption. $\endgroup$ – Sven Pistre Jan 1 '16 at 14:46
  • $\begingroup$ Not really: Because all summands are non-negative we can rearrange however we want without having to worry. This works even if the sums do not converge (then both sides are $\infty$). (This is basicly a special case of Tonelli’s theorem.) $\endgroup$ – Jendrik Stelzner Jan 1 '16 at 14:50
  • $\begingroup$ I only know Tonelli as making it possible to say that $\sum_m\sum_n a_{mn} = \sum_n\sum_m a_{mn}$. By Cauchy's convergence test we know that for convergent series $\sum_n x_n, \sum_n y_n$ we have $$\left|\sum_{n=N}^M (x_n+y_n)\right| \leq \left|\sum_{n=N}^M x_n \right| +\left|\sum_{n=N}^My_n\right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ and thus $\sum_n (x_n+y_n)$ converges. But you're right, that this does hold for divergent series with non-negative summands as well since $\sum_n x_n \leq \sum_n (x_n+y_n)$ and thereby also the right hand side diverges. $\endgroup$ – Sven Pistre Jan 1 '16 at 15:07
  • $\begingroup$ More generally Tonelli states that for $\sigma$-finite measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ and a ($\mathcal{A} \times \mathcal{B}$)-measurable function $f \colon X \times Y \to [0,\infty]$ we have $\int_{X \times Y} f(x,y) \,\text{d}(\mu\times\nu)(x,y)=\int_X\int_Y f(x,y) \,\text{d}\nu(y) \,\text{d}\mu(x)=\int_Y\int_X f(x,y) \,\text{d}\mu(x) \,\text{d}\nu(y)$. For the counting measure on $\mathbb{N}$ we get $\sum_{n,m \in \mathbb{N}} a_{mn}=\sum_{n \in \mathbb{N}}\sum_{m \in \mathbb{N}} a_{mn}=\sum_{m \in \mathbb{N}}\sum_{n \in \mathbb{N}} a_{mn}$ if $a_{nm} \geq 0$. $\endgroup$ – Jendrik Stelzner Jan 1 '16 at 15:15
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    $\begingroup$ @SvenPistre: After thinking about it for some time I figured that calling it a special case of Tonelli isn’t really such a great idea. Seems like I wasn’t as unaffected by sleepiness as I thought. This being said we can still use Tonelli to show the statement: Consider $c_{mn}$ defined as $c_{0n} = a_n^2$, $c_{1n} = b_n^2$ and $c_{mn} = 0$ for $m \geq 2$. Then $\sum_n \sum_m c_{mn} = \sum_n a_n^2 + b_n^2$ and $\sum_m \sum_n c_{mn} = \sum_n a_n^2 + \sum_n b_n^2$, both of which are the same by Tonelli. $\endgroup$ – Jendrik Stelzner Jan 2 '16 at 1:08

Hint : Use Cauchy-Schwarz inequality.$$\sum a_nb_n\le \left(\sum a_n^2\right)^{1/2}.\left(\sum b_n^2\right)^{1/2}$$

  • $\begingroup$ Can you give some counter examples $\endgroup$ – Taylor Ted Jan 1 '16 at 13:46
  • $\begingroup$ What example ?? $\endgroup$ – Empty Jan 1 '16 at 13:50
  • $\begingroup$ I want to do this question without doing proof if possible. Also can you please tell me use of Cauchy Schwartz inequality $\endgroup$ – Taylor Ted Jan 1 '16 at 13:52
  • $\begingroup$ I already mention the inequality in my answer... $\endgroup$ – Empty Jan 1 '16 at 13:53
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    $\begingroup$ What I meant is: you need the absolute values to show the absolute convergence of the series. $\endgroup$ – Clement C. Jan 1 '16 at 14:51

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