Results 1 to 5 of 5

Math Help - series convergence question

  1. #1
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1

    series convergence question

    This is probably really obvious, but I am not seeing it.

    Why does,

    \lim_{n\rightarrow \infty} (1 + \frac{1}{n})^n = e ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by mollymcf2009 View Post
    This is probably really obvious, but I am not seeing it.

    Why does,

    \lim_{n\rightarrow \infty} (1 + \frac{1}{n})^n = e ?
    Apply L'Hopital's rule. By direct evaluation, it has the indeterminate form of 1^{\infty}

    Let L=\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n

    We then see that \ln\left(L\right)=\lim_{n\to\infty}n\ln\left(1+\fr  ac{1}{n}\right)=\lim_{n\to\infty}\frac{\ln\left(1+  \frac{1}{n}\right)}{\frac{1}{n}}. Now if we directly evaluate the limit, we have the case \frac{0}{0}, which then satisfies the condition to apply L'Hopital's Rule.

    Now, \lim_{n\to\infty}\frac{\ln\left(1+\frac{1}{n}\righ  t)}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{1}{1  +\frac{1}{n}}\cdot-\frac{1}{n^2}}{-\frac{1}{n^2}}=\lim_{n\to\infty}\frac{1}{1+\frac{1  }{n}}=1

    But, \ln\left(L\right)=\lim_{n\to\infty}\frac{\ln\left(  1+\frac{1}{n}\right)}{\frac{1}{n}}=1\implies L=e^1=e

    Therefore, \boxed{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)  ^n=e}

    In a more general sense, I leave it for you to show that \boxed{\lim_{n\to\infty}\left(1+\frac{a}{n}\right)  ^n=e^a}

    Does this demystify things?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Just a little observation: the use of L'Hopital rule in demonstration that...

    \lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r} = e (1)

    ... presupposes that you have demonstrated before that...

    \frac {d}{dx} \ln x = \frac {1}{x} (2)

    ... and that presupposes you have demonstrated before that...

    \frac {d}{dx} e^{x} = e^{x} (3)

    ... and that presupposes you have demonstrated before that...

    \lim_{x \rightarrow 0} \frac{e^{x}-1}{x}= 1 (4)

    ... and that presupposes you have demonstrated before (1)...

    So invoking L'Hopital's rule for demostration of (1) is a sort of 'vicious circle' and is better to avoid it...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    you can't prove \lim_{n\to\infty}(1 + 1/n)^n=e because it's just a definition! well, unless you have another definition of e and you want to prove that the one you have implies the one that we have!

    anyway, the only thing which needs to be proved here is that \lim_{n\to\infty}(1 + 1/n)^n exists, which it does of course. now e is just a name for the limit.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    It is interesting however the demonstration that the number e can be defined as limit in two different ways...

    e= \lim_{n\rightarrow \infty} (1+\frac {1}{n})^{n} = \lim_{n\rightarrow \infty} \sum_{k=0}^{n} \frac{1}{k!}

    From the pratical point of view the accurate computation of e as...

    e = \lim_{n\rightarrow \infty} (1+\frac {1}{n})^{n}

    ... requires an n of the order of several milions, while the accurate computation of e as...

    e = \lim_{n\rightarrow \infty} \sum_{k=0}^{n} \frac{1}{k!}

    ... requires n less than 10! ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Series convergence question
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 27th 2011, 05:49 AM
  2. Replies: 3
    Last Post: November 3rd 2010, 07:55 AM
  3. Series convergence/divergence question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2010, 10:42 PM
  4. Series convergence question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 9th 2009, 06:12 AM
  5. Series convergence question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 22nd 2009, 12:55 PM

Search Tags


/mathhelpforum @mathhelpforum