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Math Help - Show that a series converges absolutely given convergence of two related series

  1. #1
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    Show that a series converges absolutely given convergence of two related series

    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?
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  2. #2
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    Re: Show that a series converges absolutely given convergence of two related series

    Quote Originally Posted by tomcruisex View Post
    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.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Show that a series converges absolutely given convergence of two related series

    Quote Originally Posted by tomcruisex View Post
    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
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    MHF Contributor chisigma's Avatar
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    Re: Show that a series converges absolutely given convergence of two related series

    Quote Originally Posted by tomcruisex View Post

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

    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?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Show that a series converges absolutely given convergence of two related series

    Quote Originally Posted by tomcruisex View Post
    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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