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Thread: P values

  1. #1
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    P values

    P values-download.png

    I need to know for which p value the serie diverges, converges and converges absolutely . Been trying to do this problem many times ;( and still dont get it. Please help.. Someone explain to me the steps because im at a loss here. I basically tried to do the alternating series test but I am confused with the ''e^-np'' as I am really bad at manipulating with e.
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    Re: P values

    Quote Originally Posted by steelmaste View Post
    Click image for larger version. 

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    I need to know for which p value the serie diverges, converges and converges absolutely . Been trying to do this problem many times ;( and still dont get it. Please help.. Someone explain to me the steps because im at a loss here. I basically tried to do the alternating series test but I am confused with the ''e^-np'' as I am really bad at manipulating with e.
    Can you find values of $p$ for which $\dfrac{x}{e^{px}(x^2-1)}$ is positive and decreasing to zero ?
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    Re: P values

    yes I can , p> equal to 0 is evidently equal to 0 and for p <0 we can do the Hopital rule right ? but it's from there that I get confused because of the E''N''P . The N infront of the P is really messing me up . Usually the problems I do just have E^p
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  4. #4
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    Re: P values

    Quote Originally Posted by steelmaste View Post
    yes I can , p> equal to 0 is evidently equal to 0 and for p <0 we can do the Hopital rule right ? but it's from there that I get confused because of the E''N''P . The N infront of the P is really messing me up . Usually the problems I do just have E^p
    Look I changed n to x in $\dfrac{x}{e^{px}(x^2-1)}$ trying to make it more familiar( recognizable).
    The alternating series test says that the positive part of the argument is always decreasing to zero.
    Again, for what values of $p$ is $f(n)= \dfrac{n}{e^{pn}(n^2-1)}$ is such that $f(n)>0~\&~ f'(n)<0~?$
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    Re: P values

    Let $\displaystyle r=e^{-p}$ and use partial fractions to expand the sum

    $\displaystyle \sum _{n=2}^{\infty } (-1)^n\frac{n}{n^2-1}r^n=\sum _{n=2}^{\infty } (-1)^n\frac{r^n}{2 (n-1)}+\sum _{n=2}^{\infty } (-1)^n\frac{r^n}{2 (n+1)}$

    we have the sum of two alternating convergent series for $\displaystyle 0<r\leq 1$

    $\displaystyle 0<e^{-p}\leq 1$

    $\displaystyle p \geq 0$
    Last edited by Idea; Oct 12th 2018 at 12:30 AM.
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