I'm confused on the limit of $\left(1+\frac{1}{n}\right)^n$

Question

Okay so I read Richard Rusczyk's AoPS Volume 2 Book, and I stumbled upon the part where he informs very briefly that $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. But he doesn't really provide a rigorous proof as to why that's true (not criticizing him or anything).. It would really help if someone could provide me with the simplest proof possible as to why $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. Thank you in advance!

Answer 1

Assume $e$ is defined by $e := \sum_\limits{k=0}^{\infty} \frac {1}{k!}$

Let

$$y=\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$ $$\ln y=\lim_{n \rightarrow \infty} n\ln\left(1+\frac{1}{n}\right)$$ $$\ln y=\lim_{n \rightarrow \infty} \frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$ using L'Hôpital's rule $$\ln y = \lim_{n \rightarrow \infty} \frac{\frac{-1}{n^{2}}*\frac{1}{1+\frac{1}{n}}}{\frac{-1}{n^{2}}}$$ $$\ln y = \lim_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}$$ $$\ln y = 1$$ $$y=e$$

Answer 2

Using binomial expansion

$$(1+1/n)^n=1+n(1/n)+n(n-1)/2!(1/ n)^2+...$$

$$=1+(1/1!)+(1/2!)(1-\frac{1}{n})+...$$

As $n\rightarrow\infty$, RHS$=1+(1/1!)+(1/2!)+...=e$

Answer 3

What is your definition of $e?$

In many books by definition $e = \lim_\limits{n\to\infty} (1+\frac 1n)^n.$

In some books $e$ is defined as $e = \sum_\limits{k=0}^{\infty} \frac {1}{k!}$

But we can prove these are equal.

Lets define $e = \sum_\limits{k=0}^{\infty} \frac {1}{k!}$

This proof can be found in baby Rudin.

Let $t_n = (1+\frac 1n)^n$

and $s_n = \sum_\limits{k=0}^{n} \frac {1}{k!}$

Expand $(1+\frac {1}{n})^n$ using the binomial theorem.

$t_n = 1 + \frac {n}{n} + \frac 12 (1 - \frac 1n) + \frac 1{3!} (1 - \frac 1n)(1-\frac 2n) + \cdots + \frac {1}{n!} (1-\frac {1}{n})\cdots(1-\frac {n-1}{n})$

Every term in $t_n$ is less than or equal to its corresponding term in $s_n$

$t_n \le s_n$

$\limsup_\limits{n\to \infty} t_n \le e$

Choose some $m < n$

Looking at the first $m$ terms of $t_n$

$1 + 1 + \frac 12 (1 - \frac 1n) + \frac 1{3!} (1 - \frac 1n)(1-\frac 2n) + \cdots + \frac {1}{m!} (1-\frac {1}{n})\cdots(1-\frac {m-1}{n})$

Let $n$ approach infinity, keeping $m$ fixed.

$1 + 1 + \frac 1{2} + \frac 1{3!} + \cdots + \frac {1}{m!} = s_m$

Since there are additional terms in $\lim_\limits{n\to\infty} (1+\frac 1n)^n$ and all are positive.

$s_m \le \liminf_\limits{n\to\infty} (1+\frac 1n)^n$

Letting $m$ go to infinity.

$e\le \liminf_\limits{n\to\infty} (1+\frac 1n)^n$

And by the squeeze theorem, $\lim_\limits{n\to\infty} (1+\frac 1n)^n = e$

Answer 4

This is of the form $1^\infty$ as $ n \to \infty$.

Now for any limit $\displaystyle\lim_{n\to \infty}f(x)^{g(x)}$ of the form $1^\infty$ we have direct formula to be : $e^{lim_{n\to\infty}(f(x)-1)(g(x))}$.

Now $f(x)=1+\frac{1}{n}$ and $g(x)=n$. Evaluating using the above formula the limit comes out to be : $$e^{lim_{n\to \infty}(1+\frac{1}{n}-1)n}=e^{lim_{n \to \infty}(\frac{1}{n})n}=e.$$

Answer 5

I think at that stage all you need to know is that the limit exists. We simply call it $e,$ for short.

So how can we tell that the limit exists?

The simplest way I know is already provided as an answer (albeit partial; this answer completes it) -- expand the binomial, and when you take limits as $n\to \infty,$ you get an infinite series which can be proven to exist by elementary comparisons, thus.

So we have $$1+1+1/2+1/3!+1/4!+1/5!+\cdots\le 1+1+1/2+1/2^2+1/2^3+1/2^4+\cdots=1+1/(1-1/2)=1+2=3.$$ Since the series has only positive terms, the sequence of partial sums increases monotonically, and therefore the limit must exist. We've simultaneously shown that this number $e$ is not up to $3.$ In fact, by adding up sufficiently many terms of the series, we obtain $e=2.718281828459045\ldots,$ to the first $15$ places of decimal. Stare at it -- it's easy to memorise.