Results 1 to 5 of 5

Thread: 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,

    $\displaystyle \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
    5
    Quote Originally Posted by mollymcf2009 View Post
    This is probably really obvious, but I am not seeing it.

    Why does,

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

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

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

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

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

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

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

    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
    6
    Just a little observation: the use of L'Hopital rule in demonstration that...

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

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

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

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

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

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

    $\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x}= 1$ (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

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

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    you can't prove $\displaystyle \lim_{n\to\infty}(1 + 1/n)^n=e$ because it's just a definition! well, unless you have another definition of $\displaystyle e$ 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 $\displaystyle \lim_{n\to\infty}(1 + 1/n)^n$ exists, which it does of course. now $\displaystyle e$ 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
    6
    It is interesting however the demonstration that the number e can be defined as limit in two different ways...

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    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: Jan 27th 2011, 04:49 AM
  2. Replies: 3
    Last Post: Nov 3rd 2010, 06:55 AM
  3. Series convergence/divergence question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2010, 09:42 PM
  4. Series convergence question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 9th 2009, 05:12 AM
  5. Series convergence question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Mar 22nd 2009, 11:55 AM

Search Tags


/mathhelpforum @mathhelpforum