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How to prove the following limit?

$$\lim_{n \to \infty} (1+1/n)^n = e$$

I can only observe that the limit should be a very large number!

Thanks.

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Just apply the binomial theorem then move things around a bit: $$\left(1+\tfrac{1}{n}\right)^n = \sum_{k=0}^n \frac{{n \choose k}}{n^k} = \sum_{k=0}^n \frac{1}{k!} \frac{n!/(n-k)!}{n^k}.$$

Now if you could just get rid of that $\frac{n!/(n-k)!}{n^k}$ term… (using the idea of $n \to \infty$ and $k$ is usually small in comparison)

**edit** Jonas makes a good point, I was implicity assuming the definition $e = \sum_{k = 0}^\infty \frac{1}{k!}$

Actually, the way things work out in mathematics usually is that we only prove that $x_n = (1+1/n)^n$ is a convergent sequence, and we *define* its limit to be $e$. Some people use other definitions for $e$ and show that it is equivalent to this definition, but there are many ways to do this that are logically equivalent.

*(Just for the record, you don’t “prove” limits, you compute them.)*

One way to show that this sequence is convergent is to show that it is increasing and bounded above. This is a bit technical but works out quite well. To prove that $x_n$ is increasing, we show that $x_{n+1} \ge x_n$, which is equivalent to $x_{n+1}/x_n \ge 1$ :

\begin{align}

\frac{x_{n+1}}{x_n} & = \frac{ \left( 1 + \frac 1{n+1} \right)^{n+1} }{\left( 1 + \frac 1n \right)^n } = \left( 1 + \frac 1{n+1} \right) \left( \frac{ 1 + \frac 1{n+1} }{1 + \frac 1n} \right)^n = \left( \frac {n+2}{n+1} \right) \left( \frac{(n+2)n}{(n+1)^2} \right)^n \\

& = \left( \frac{n+2}{n+1} \right) \left( 1 + \frac{(n+2)n – (n+1)^2 }{(n+1)^2} \right)^n \\

& = \left( \frac{n+2}{n+1} \right) \left( 1 – \frac{1}{(n+1)^2} \right)^n \\

& \ge \left( \frac{n+2}{n+1} \right) \left( 1 – \frac n{(n+1)^2} \right) \\

& \ge \left( \frac{n+2}{n+1} \right) \left( \frac{n^2 + n + 1}{n^2 + 2n + 1} \right) = \frac{n^3 + 3n^2 + 3n + 2}{n^3 + 3n^2 + 3n + 1} \ge 1.

\end{align}

The first inequality holds by *Bernoulli’s inequality* ( I’m referring to $(1+x)^n \ge 1+nx$ for $x \ge -1$).

To prove that $x_n$ is bounded above, we use the binomial theorem :

\begin{align}

\left(1+\frac 1n\right)^n & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac 1{n^k} = \sum_{k=0}^n \frac 1{k!} \frac{n!}{n^k(n-k)!} \\

& \le 1 + \sum_{k=1}^n \frac 1{k!} \le 1 + \sum_{k=1}^n \frac 1{2^{k-1}} = 1 + \frac{1- \frac 1{2^n}}{ 1- \frac 12} < 1 + 2 = 3. \\

\end{align}

Note that I used the fact that $k! \ge 2^{k-1}$, which implies $\frac 1{2^{k-1}} \ge \frac 1{k!}$. (To see this, just write $k! = k \times \dots \times 2 \times 1 \ge 2 \times \dots \times 2 \times 1 = 2^{k-1}$.)

Another way would be to show that this sequence converges to $\sum_{k=0}^{\infty} \frac 1{k!}$ (which is another definition for $e$) by expanding $(1+1/n)^n$ with the binomial theorem.

Hope that helps,

The fact that it converges to a limit is already explained in other answers. Then to get an idea as to why this limit must match the definition of $e$:

We know that $\exp(x) = e^x$ is a function that starts with $\exp(0) = 1$, and grows with a rate that is equal to it’s current value (ie, $\frac{d}{dx}e^x = e^x$). So a good way to find this, is to approximate it with a finite number of segments, where the rate at each segment is constant, and equal to the value at the beginning of the segment:

$\exp_n(0) = 1$

$\exp_n(t) = \exp_n(t – \frac{1}{n}) + \frac{1}{n}\exp_n(t – \frac{1}{n})$

$\exp_n(t) = (1 + \frac{1}{n})\exp_n(t – \frac{1}{n})$

Clearly, taking smaller and smaller segments will give a better approximation of the actual smooth function, with a constantly changing rate, and in the limit, it will match the definition of $\exp$. Unrolling the recursion for $\exp_n(1)$ clearly gives

$\exp_n(1) = (1 + \frac{1}{n})^n \exp_n(0) = (1 + \frac{1}{n})^n$

so

$\exp(1) = \lim_{n \rightarrow \infty} \exp_n(1) = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$

This is enough to prove that if the limit exists, then it matches the definition of $e$.

One way to prove it is to manipulate the the exponential of the logarithm. Note that we can write:

\begin{equation*}

\lim_{n\to\infty} e^{n\ln(1+\frac{1}{n})}=e^{\lim_{n\to\infty}n\ln (1+\frac{1}{n})}=e^{\lim_{n\to\infty}\frac{\ln(1+\frac{1}{n})}{1/n}}.

\end{equation*}

An application of L’Hopital’s rule gives

\begin{equation*}

e^{\lim_{n\to\infty}\frac{1}{1+1/n}}=e^1=e.~_{\square}

\end{equation*}

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