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

Why does,

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

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- Apr 14th 2009, 07:49 PMmollymcf2009series 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 $ ? - Apr 14th 2009, 08:34 PMChris L T521
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? - Apr 14th 2009, 10:36 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 14th 2009, 11:43 PMNonCommAlg
you can't prove $\displaystyle \lim_{n\to\infty}(1 + 1/n)^n=e$ because it's just a definition! (Giggle) 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! (Nod)

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. - Apr 15th 2009, 12:23 AMchisigma
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! (Wink)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$