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Math Help - [SOLVED] Two Series for Convergence

  1. #1
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    [SOLVED] Two Series for Convergence

    Problem
    Examine the following series for convergence
    1. \sum_{k=1}^{\infty}ke^{-k^2}
    2. \sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3
    =================================
    Attempt:
    1. This series \sum_{k=1}^{\infty}ke^{-k^2} can be rewritten as \sum_{k=1}^{\infty} \frac{k}{e^{k^2}}.
    I try to use the Ratio Test. By the Ratio Test

    \lim_{k \to \infty} \left| \frac{k+1}{e^{(k+1)^2}} \frac{e^{k^2}}{k}\right| = \lim_{k \to \infty} \left| \frac{e^{k^2}}{e^{(k+1)^2}} \left( 1 + \frac{1}{k}\right) \right|

    I know the term 1 + 1/k goes to 1 as  k \rightarrow \infty . However I do know about the left term. \frac{e^{k^2}}{e^{(k+1)^2}}

    2. I tried to find if this series converges or not by the Comparison Test but went nowhere.

    Thank you for your time.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Paperwings View Post
    Problem
    Examine the following series for convergence
    1. \sum_{k=1}^{\infty}ke^{-k^2}
    2. \sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3
    =================================
    Attempt:
    1. This series \sum_{k=1}^{\infty}ke^{-k^2} can be rewritten as \sum_{k=1}^{\infty} \frac{k}{e^{k^2}}.
    I try to use the Ratio Test. By the Ratio Test

    \lim_{k \to \infty} \left| \frac{k+1}{e^{(k+1)^2}} \frac{e^{k^2}}{k}\right| = \lim_{k \to \infty} \left| \frac{e^{k^2}}{e^{(k+1)^2}} \left( 1 + \frac{1}{k}\right) \right|

    I know the term 1 + 1/k goes to 1 as  k \rightarrow \infty . However I do know about the left term. \frac{e^{k^2}}{e^{(k+1)^2}}

    2. I tried to find if this series converges or not by the Comparison Test but went nowhere.

    Thank you for your time.
    \frac{e^{k^2}}{e^{(k+1)^2}} = \frac{e^{k^2}}{e^{k^2 + 2k + 1}} = \frac{e^{k^2}}{e^{k^2}e^{2k}e} \longrightarrow ...
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Paperwings View Post
    2. \sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3
    =================================
    \left(\frac{k+1}{k^2+1}\right)^3 < \left(\frac{k+1}{k^2}\right)^3 = \frac{k^3 + 3k^2 + 3k + 1}{k^6} ...
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  4. #4
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    Quote Originally Posted by kalagota View Post
    \frac{e^{k^2}}{e^{(k+1)^2}} = \frac{e^{k^2}}{e^{k^2 + 2k + 1}} = \frac{e^{k^2}}{e^{k^2}e^{2k}e} \longrightarrow ...
    Then \frac{e^{k^2}}{e^{k^2}e^{2k}e} = \frac{1}{e^{2k+1}}\rightarrow 0
    Since 0 <1, then the series converges absolutely by the Ratio Test.

    For the second one, from your hint
    \frac{k^3 + 3k^2 + 3k + 1}{k^6} = \frac{1}{k^3}+\frac{3}{k^4}+\frac{3}{k^5}+\frac{1}  {k^6}. Then by the p-Test ( the series \sum_{k=1}^{\infty} \frac{1}{k^p} converges if and only if p > 1) since all k's are > 1 then the series converge.

    Thank you Kalagota.
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