How can we solve $\lim\limits_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$?

Question

The number $e$ is the limit $$e = \lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$

Graph of $f(x) = \left (1 + \dfrac{1}{x} \right)^x$ taken from here.

Its evident from the graph that the limit actually approaches $e$ as $x$ approaches $\infty$. So I tried approaching the value algebraically. My attempt:

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

which is an indeterminate form. I cannot think of any other algebraic manipulation. My question is that how can I solve this limit algebraically?

Usually, we prove that the sequence given by $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing and bounded, hence convergent, and just call $e$ its limit. With this assumption, there is nothing to prove. — Jack D'Aurizio, Oct 15 '16 at 19:27
Perhaps have a look at this post and other questions linked there — Martin Sleziak, Oct 16 '16 at 09:07

score 5 · Accepted Answer · answered Oct 15 '16 at 19:10

An algebraic way is Binomial Expansion, which is given by $$\begin{eqnarray*} \left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\ &=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot \frac{1}{3!}+\cdots \\ \\ \end{eqnarray*}$$ Now every term in front there goes to $1$ for $n\to \infty$, so that the limit gives $$ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{\!n} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots = \mathrm{e}. $$

score 0 · Answer 2 · answered Oct 15 '16 at 19:08

0

Notice that

$$\left(1+\dfrac{1}{n}\right)^n=e^{n\ln\left(1+\frac{1}{n}\right)}$$

Can you solve the limit now?

answered Oct 15 '16 at 19:08

It appears strange to me. Do we need to use $e$ to find the value of $e$? – Parth Oct 15 '16 at 19:09
I don't know any way of solving it without prior knowledge of $e$. Maybe the other answer is more satisfying! – Oct 15 '16 at 19:11
@unknownCoder not really. You can write $(1+\frac{1}{n})^n = 2^{n \log_2(1+1/n)}$, and you'll have to prove that $\lim_{n \to \infty} n \log_2(1+1/n) = 1/\ln(2)$, i.e. that the derivative of $\log_2(1+x)$ is $\frac{1 / \ln(2)}{1+x}$ – reuns Oct 15 '16 at 19:12
1

@unknownCoder note this is a definition of $e$ : $\log_a' (x) = \frac{f(a)}{x}$ where $1/f(a)$ is a logarithm in some base that we'll call $e$, such that $\log_e'(x) = \frac{1}{x}$ – reuns Oct 15 '16 at 19:16

How can we solve $\lim\limits_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$?

2 Answers2