Results 1 to 3 of 3

Thread: Log(x+1)

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    7

    Log(x+1)

    $\displaystyle \int^x_0\sum^\infty_{n=1}\frac{(-1)^{n-1}t^{n}}{n}=\int^x_0\sum^\infty_{n=1}(-1)^{n-1}t^{n-1}dt\underbrace{=}_{\mbox{justify}}\int^x_0\frac{1 }{1+t}dt=\log(1+x)$

    Ok so I need to prove that the middle equality it true, but I don't have the slightest clue where to begin.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Belgium
    Posts
    1,254
    Thanks
    24

    Re: Log(x+1)

    You have to prove that:
    $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}t^{n-1}=\frac{1}{1+t}$
    It can be useful to write a few terms of the summation:
    $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}t^{n-1}=1-t+t^2-t^3+...+(-1)^{k-1}t^{k-1}+...$

    Do you notice a 'special' serie? Can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Log(x+1)

    $\displaystyle 1-t+t^2-t^3+t^4-....$ is an infinite geometric series with $\displaystyle a=1$ and $\displaystyle r=-t.$

    Therefore,$\displaystyle 1-t+t^2-t^3+t^4-.... = \frac{a}{1-r}$
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum