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$e$ - Euler's constant

$e = \lim_{n\to\infty}(1 + \frac1n)^n \approx 2.71828$

I'm wondering how 2.71828 is obtained from that.

user1766555
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  • Have you looked up the various definitions of Euler's constant? http://en.wikipedia.org/wiki/E_(mathematical_constant) – Ian Coley Sep 08 '13 at 21:58
  • I'm wondering what mathematical operations take place that give the number however. – user1766555 Sep 08 '13 at 22:02
  • Addition and multiplication? If you take the equivalent definition $e=\sum_{n=0}^\infty \frac{1}{n!}$ maybe that will satisfy you better. – Ian Coley Sep 08 '13 at 22:03
  • In my answer here you can see how to find $e$ to one decimal digit. Take more terms and you can narrow the bounds and get as many digits as you like. –  Sep 08 '13 at 22:04
  • But doesn't $ \frac1n $ become $0$ in this? Also I'm not sure about the exponent n here. – user1766555 Sep 08 '13 at 22:09
  • Usually, $e \approx 2.718$ is Euler's number, while $\gamma \approx 0.577$ is Euler's constant. – Argon Sep 08 '13 at 23:35

3 Answers3

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In this answer, it is shown that if we define $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{1} $$ then $$ e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{2} $$ Summing the series in $(2)$ converges pretty quickly for $x=1$ to give $$ e=2.718281828459\dots\tag{3} $$ If you wish to show that the limit in $(1)$ exists, this answer shows that $$ \left(1+\frac1n\right)^n\tag{4} $$ is an increasing sequence and that $$ \left(1+\frac1n\right)^{n+1}\tag{5} $$ is a decreasing sequence. Thus, the limit of both lies between each.

Community
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robjohn
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The number $e$ has an infinite, non-terminating, non-repeating decimal expansion, $2.71828$ is $e$ truncated to $6$ significant figures. As you increase $n$, $$\left(1 + \frac{1}{n}\right)^{n} \to e$$

This means that you can make the difference between $e$ and $\left(1 + \frac{1}{n}\right)^{n} $ as small as you'd like by increasing $n$.

Rustyn
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  • Doesn't the whole thing turn into $ \infty $? And doesn't $ \frac1n $turn into $0$ ? – user1766555 Sep 08 '13 at 22:26
  • @user1766555 No, after "evaluating" it near infinity you have $(1^\infty)$. Its indeterminate, like $\frac{0}{0}$, or $\frac{\infty}{\infty}$. It needs further simplification before a result can be obtained. You are right that 1/n approaches 0, leaving us with $1^{\infty}$ – ra1nmaster Sep 08 '13 at 22:27
  • So how is 2.71828 determined from an indeterminate? – user1766555 Sep 08 '13 at 22:33
  • @user1766555 You do simplification work on the limit. See my answer below, I raised e to the power of the natural log of $(1 + 1/n)^n$, keeping the value the same. This limit was easier to work on. – ra1nmaster Sep 08 '13 at 23:01
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Edit: First, try looking at the values of $f(x) = (1 + 1/x)^x$ for large x. You will see that as they get larger, they seem to level off at a value. That it is, as x approaches infinity, the limit approaches a certain value. You should notice that value to be around the number you mentioned. The larger you make $x$, the closer to $e$ you get, or numerically 2.71828...

Do you want a method for calculating the limit definition to see how it equals $e$? I guess you could do it like this:

$\lim_{x\rightarrow \infty}\left ( 1+\frac{1}{x} \right )^x=\lim_{x\rightarrow \infty}\ e^{\ln\left ( {\left ( 1+\frac{1}{x} \right )^x} \right )}$.

because e to the power of the natural log of some quantity is just that quantity, as $e^x$ and $\ln(x)$ are inverses.

Using the property that $\ln(a^{b}) = b\ln{a}$:

$\lim_{x\rightarrow \infty}\ e^{\ln\left ( {\left ( 1+\frac{1}{x} \right )^x} \right )} =\lim_{x\rightarrow \infty}\ e^{x\ln\left ( {\left ( 1+\frac{1}{x} \right )} \right )}$.

Notice that we can rewrite the above like this:

$\lim_{x\rightarrow \infty}\ e^{x\ln\left ( {\left ( 1+\frac{1}{x} \right )} \right )}$=$e^{\lim_{x\rightarrow \infty}\ x \ln (1 + \frac{1}{x})} $. We brought the limit into the exponent. The limit $\lim_{x\rightarrow \infty} x\ln(1 + \frac{1}{x})$ happens to equal one (see below for explanation), and therefore the exponent on the $e$ is one, and $e^1$ = $e$. Therefore, the limit is $e^1 = e$ Here's why the the limit in the exponent is 1:

The exponent on the e can be determined by $\lim_{x\rightarrow \infty} x\ln(1 + \frac{1}{x})$ , as shown above. To use calculate it, we may let $u = \frac{1}{x}$, and therefore, as $x\rightarrow \infty$, $u\rightarrow 0$. So, substituting $u = \frac{1}{x}$ in our limit and changing $x\rightarrow \infty$ to $u\rightarrow 0$:

$\lim_{x\rightarrow \infty} x\ln(1 + \frac{1}{x})=\lim_{u \rightarrow 0} \frac{1}{u}\ln(1 + u)$. The denominator and numerator go to $0$, so by L'Hopital's:

$\lim_{u \rightarrow 0} \frac{1}{u}\ln(1 + u)=\lim_{u \rightarrow 0} \frac{(\ln(1 + u))'}{u'}$=$\lim_{u \rightarrow 0} \frac{(1/(1 + u))}{1}$ = $1$.

So the whole limit is $e^1 = e$.

ra1nmaster
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