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Integral Test of e^n/(e^(2n)+9)

  1. May 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Using the integral test determine whether or not the following sum diverges. $$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9} $$

    2. Relevant equations
    3. The attempt at a solution
    $$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9}=\sum_{n=1}^{\infty} \frac{e^n}{({e^{n})}^2+9} \\
    \int^{\infty} \frac{e^n}{({e^{n})}^2+9} \cdot dn; \hspace{0.5cm} Let \space u=e^n; \hspace{0.5cm} du=e^n\cdot dn \\
    \int^{\infty} \frac{du}{u^2+9}; \hspace{0.5cm} Let \space u=3\tan\theta \hspace{0.5cm} du=3\sec^2\theta \\
    \int^{(\frac{\pi}{2})^-} \frac{3\sec^2\theta}{9\tan^2\theta+9}=\frac{1}{3} \int^{(\frac{\pi}{2})^-} 1\cdot d\theta=\frac{\pi}{6} $$ Since I didn't get infinity in my answer this sum converges however my answer key says that it does diverge, did I miss something?
     
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  3. May 20, 2015 #2

    SammyS

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    What is the lower limit for the integral?

    It is true that this series converges.

    Added in Edit:

    You have left dθ out of a couple of expressions.
     
    Last edited: May 21, 2015
  4. May 20, 2015 #3
    I believe it would be ##arctan(\frac{1}{3}) ## although my textbook says "The integral is to be evaluated only at the upper limit; no lower limit is needed" Edit: Well this is embarrassing I was looking at the answer for 5.10 instead of 6.10, whoops
     
  5. May 21, 2015 #4

    SammyS

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    If you write an integral as a definite integral, it needs both limits.

    If you evaluate the indefinite integral at the limit corresponding to infinity, that should be adequate, as your textbook says.

    I belive the lower limit for your last integral is ##\displaystyle \ arctan\left(\frac{e}{3}\right) ##
     
  6. May 21, 2015 #5
    So is the notation they're using in my textbook wrong then? Capture.jpg
     
  7. May 21, 2015 #6

    SammyS

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    In my opikion, that's not a standard notation.

    If your textbook defines it, then it's fine to use in the context of your textbook & class, but I'm not familiar with the notation.
     
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