Show that a series converges absolutely given convergence of two related series

**tomcruisex** · October 2nd 2011, 11:37 PM

Show that if $\sum_{n=1}^{\infty}a_n^2$ and $\sum_{n=1}^{\infty}b_n^2$ are convergent then $\sum_{n=1}^{\infty}a_nb_n$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$|xy|\leq 1/2(|x|^2 + |y|^2)$
$2|xy| \leq |x|^2 + |y|^2$
$|x|^2 + |y|^2 - 2|xy| \geq 0$
$(|x| - |y|)^2 \geq 0$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?

**Prove It** · October 2nd 2011, 11:57 PM

Originally Posted by tomcruisex

Show that if $\sum_{n=1}^{\infty}a_n^2$ and $\sum_{n=1}^{\infty}b_n^2$ are convergent then $\sum_{n=1}^{\infty}a_nb_n$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$|xy|\leq 1/2(|x|^2 + |y|^2)$
$2|xy| \leq |x|^2 + |y|^2$
$|x|^2 + |y|^2 - 2|xy| \geq 0$
$(|x| - |y|)^2 \geq 0$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?

First notice that $\displaystyle |x|^2 = x^2$ and $\displaystyle |y|^2 = y^2$ .

Since you have shown that $\displaystyle |xy| \leq \frac{1}{2}\left(x^2 + y^2\right) = \frac{1}{2}x^2 + \frac{1}{2}y^2$ , that means

$\displaystyle \sum{|xy|} \leq \frac{1}{2}\sum{x^2} + \frac{1}{2}\sum{y^2}$ .

Since you know that the two sums on the right are convergent, their sum is also convergent, and anything less than that is also convergent.

Q.E.D.

**chisigma** · October 3rd 2011, 12:01 AM

Originally Posted by tomcruisex

Show that if $\sum_{n=1}^{\infty}a_n^2$ and $\sum_{n=1}^{\infty}b_n^2$ are convergent then $\sum_{n=1}^{\infty}a_nb_n$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

I can prove the Hint -
$|xy|\leq 1/2(|x|^2 + |y|^2)$
$2|xy| \leq |x|^2 + |y|^2$
$|x|^2 + |y|^2 - 2|xy| \geq 0$
$(|x| - |y|)^2 \geq 0$

And this is true for all x, y as the square of any real number will be greater than or equal to 0.

But I cant see how to apply this to the original question?

Setting $c_{n}^{2}= \text{max}\ [a_{n}^{2}\ ,\ b_{n}^{2}]$ it is easy to demonstrate that the series $\sum_{n=1}^{\infty} c_{n}^{2}$ converges. But is $|a_{n}\ b_{n}|\le c_{n}^{2}$ so that the series $\sum_{n=1}^{\infty} |a_{n}\ b_{n}|$ also converges...

Kind regards

$\chi$ $\sigma$

**chisigma** · October 3rd 2011, 12:25 AM

Originally Posted by tomcruisex

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R...

If $|x\ y|\le \frac{|x|^{2} + |y|^{2}}{2}$ , then ...

$x^{2}\ y^{2} \le \frac{x^{4}+2 x^{2} y^{2} + y^{4}}{4} \implies \frac{x^{2} y^{2}}{2} \le \frac{x^{4}+ y^{4}}{4} \implies x^{4}+y^{4} \ge 2 x^{2} y^{2}$ (1)

Very well!... but what does it happen if $x=y=\frac{1}{100}$ ?...

Kind regards

$\chi$ $\sigma$

**tomcruisex** · October 3rd 2011, 12:51 AM

Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of $\sum_{n=1}^{\infty}a_n^2$ ?
Does that mean that the series is made up by squaring each term and adding them together - a0a0 + a1a1 + a2a2 + ... + anan?
Or does it mean that the series you get the sum of series $\sum_{n=1}^{\infty}a_n$ and then square it?

**chisigma** · October 3rd 2011, 01:12 AM

Originally Posted by tomcruisex

Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of $\sum_{n=1}^{\infty}a_n^2$ ?
Does that mean that the series is made up by squaring each term and adding them together - a0a0 + a1a1 + a2a2 + ... + anan?
Or does it mean that the series you get the sum of series $\sum_{n=1}^{\infty}a_n$ and then square it?

The meaning of $\sum_{n=1}^{\infty} a^{2}_{n}$ is [probably...] that You have a series with all positive terms. The reason of that is easily understandable considering the case $a_{n}=b_{n}= \frac{(-1)^{n}}{\sqrt{n}}$ , where both the series $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ converge but the series $\sum_{n=1}^{\infty}a_{n} b_{n}$ diverges...

Kind regards

$\chi$ $\sigma$

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