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Math Help - Convergence test for two series

  1. #1
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    Convergence test for two series

    Hello,

    I am having trouble finding out if the following two series converge.

    a) \sum_{n=1}^{+\infty}{\frac{n+1}{\sqrt{n}+1}\sin{\f  rac{1}{n^2}}}

    b) \sum_{n=1}^{+\infty}{\log\left({1+\frac{1}{n^2}}\r  ight)}

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by thomasdotnet View Post
    Hello,

    I am having trouble finding out if the following two series converge.

    a) \sum_{n=1}^{+\infty}{\frac{n+1}{\sqrt{n}+1}\sin{\f  rac{1}{n^2}}}

    b) \sum_{n=1}^{+\infty}{\log\left({1+\frac{1}{n^2}}\r  ight)}

    Any help would be appreciated.
    Compare the first with

     <br />
\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n}}<br />

    and the second with

     <br />
\sum_{n=1}^{\infty} \frac{1}{n^2}<br />
.
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  3. #3
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    For the second one, you can also write  log(1 + \frac{1}{n^2}) = \int_1^{1 + \frac{1}{n^2} } \frac{du}{u} .

    By the MVT of integral calculus we have  \int_1^{1 + \frac{1}{n^2} } \frac{du}{u} = \frac{1}{n^2} \frac{1}{\xi} , where  1 \le \xi \le 1 + \frac{1}{n^2} \Leftrightarrow \frac{1}{\xi} \le 1 , and so:

     log(1 + \frac{1}{n^2}) = \frac{1}{n^2} \frac{1}{\xi} \le \frac{1}{n^2} .

    What can you say about the convergence/divergence of the series now?
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  4. #4
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    Thank you very much guys for your help! The second one clearly converges (the integral explanation helped me a lot in understanding why I need to compare this to \frac{1}{n^2}).

    One thing I cannot completely understand is how you got to \frac{1}{n\sqrt{n}} for the first series.
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  5. #5
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    Quote Originally Posted by thomasdotnet View Post
    Thank you very much guys for your help! The second one clearly converges (the integral explanation helped me a lot in understanding why I need to compare this to \frac{1}{n^2}).

    One thing I cannot completely understand is how you got to \frac{1}{n\sqrt{n}} for the first series.
    Consider  <br />
\frac{n+1}{\sqrt{n}+1}\sin \frac{1}{n^2}<br />

    Since \lim_{n \to \infty} \frac{\sin \frac{1}{n^2}}{\frac{1}{n^2}} = 1\;\;\;

    and for n large  <br />
\frac{n+1}{\sqrt{n}+1} \approx \sqrt{n} then we compare with

     <br />
\sqrt{n} \frac{1}{n^2} = \frac{1}{n \sqrt{n}} .
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  6. #6
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    OK, that makes it clearer. Thank you!
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