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

Oct 2011
2
0
Show that if \(\displaystyle \sum_{n=1}^{\infty}a_n^2\) and \(\displaystyle \sum_{n=1}^{\infty}b_n^2\) are convergent then \(\displaystyle \sum_{n=1}^{\infty}a_nb_n\) is absolutely convergent.

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


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

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

MHF Helper
Aug 2008
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Show that if \(\displaystyle \sum_{n=1}^{\infty}a_n^2\) and \(\displaystyle \sum_{n=1}^{\infty}b_n^2\) are convergent then \(\displaystyle \sum_{n=1}^{\infty}a_nb_n\) is absolutely convergent.

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


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

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 \(\displaystyle \displaystyle |x|^2 = x^2\) and \(\displaystyle \displaystyle |y|^2 = y^2\).

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

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

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.
 
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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Show that if \(\displaystyle \sum_{n=1}^{\infty}a_n^2\) and \(\displaystyle \sum_{n=1}^{\infty}b_n^2\) are convergent then \(\displaystyle \sum_{n=1}^{\infty}a_nb_n\) is absolutely convergent.

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


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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R...
If \(\displaystyle |x\ y|\le \frac{|x|^{2} + |y|^{2}}{2}\), then ...

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Oct 2011
2
0
Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of \(\displaystyle \sum_{n=1}^{\infty}a_n^2\)?
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 \(\displaystyle \sum_{n=1}^{\infty}a_n\) and then square it?
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Thank you for the help guys.

There is one thing I am still not sure about - what is the meaning of \(\displaystyle \sum_{n=1}^{\infty}a_n^2\)?
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 \(\displaystyle \sum_{n=1}^{\infty}a_n\) and then square it?
The meaning of \(\displaystyle \sum_{n=1}^{\infty} a^{2}_{n}\) is [probably...] that You have a series with all positive terms. The reason of that is easily understandable considering the case \(\displaystyle a_{n}=b_{n}= \frac{(-1)^{n}}{\sqrt{n}}\), where both the series \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) and \(\displaystyle \sum_{n=1}^{\infty}b_{n}\) converge but the series \(\displaystyle \sum_{n=1}^{\infty}a_{n} b_{n}\) diverges...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)