# Famous Limit

$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n,\quad n\in\mathbb{N}$ is a famous limit that is in all Calc I books. Here are some proofs about it that are usually done other ways.

Theorem IA

$\left(1+\frac{1}{n}\right)^n$ is increasing

Proof:
A method Polytope saw someplace.

Consider the $n+1\,$ numbers $\left[1,\ 1+\frac{1}{n},\ 1+\frac{1}{n},\ \cdots ,\ 1+\frac{1}{n}\right]$

These have a geometric mean. $G = \left( 1\, \left(1+\frac{1}{n}\right)\, \cdots\, \left(1+\frac{1}{n}\right)\right)^{\frac{1}{n+1}}=\left(1+\frac{1}{n}\right)^{\frac{n}{n+1}}$

These have an arithmetic mean.
$A = \frac{1}{n+1}\left( 1\,+\, \left(1+\frac{1}{n}\right)\,+ \cdots\, +\left(1+\frac{1}{n}\right)\right)=1+\frac{1}{n+1}$

Now we use the Arithmetic Mean-Geometric Mean Inequality (http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means). $A \ge M$ and $A^{n+1} \ge M^{n+1}$

$\left(1+\frac{1}{n+1}\right)^{n+1} \ge \left(1+\frac{1}{n}\right)^n \qquad \square$

Theorem IB

$\left(1+\frac{1}{n}\right)^{n+1}$ is decreasing.

The proof of IA can be adapted to prove IB.

Consider the $n+2\,$ numbers $\left[1,\ 1+\frac{1}{n},\ 1+\frac{1}{n},\ \cdots ,\ 1+\frac{1}{n}\right]$

These have a geometric mean. $G = \left( 1\, \left(1+\frac{1}{n}\right)\, \cdots\, \left(1+\frac{1}{n}\right)\right)^{\frac{1}{n+2}}= \left(1+\frac{1}{n}\right)^{\frac{n+1}{n+2}}$

Next we can compare $G\,$ to the harmonic mean, $H\,$, using the Geometric Mean - Harmonic mean inequality. (http://planetmath.org/encyclopedia/ArithmeticGeometricMeansInequality.html)

$H=\frac{n+2}{1+\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{1}{n}} + \cdots + \frac{1}{1+\frac{1}{n}}}=\frac{n+2}{n+1}=1+\frac{1}{n+1}$

$G= \left(1+\frac{1}{n}\right)^{\frac{n+1}{n+2}} \ge H=1+\frac{1}{n+1}$

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

$\square$

Theorem IC

The sequences from IA and IB both converge to the same limit.

Proof

Since the sequence of IB is decreasing and positive it is bounded below by 0. The sequence of IB converges due to Proposition 3.1.9 at this site. (http://pirate.shu.edu/~wachsmut/ira/numseq/sequence.html)

For all n, $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{n}\right)^{n+1} \le \left(1+\frac{1}{1}\right)^{1+1}=4$

So the convergence of the sequence from IA follows by the same proposition.

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

So both sequences have the same limit.

$\square$

Lemma IIL

$\lim_{x\rightarrow \infty} \int_1^{x} \frac{dt}{t} = \infty$

Proof

As $x\,$ increases, the integral smoothly and monotonically increases beause its first derivative is $\frac{1}{x}$. Suppose it reaches a finite limit $L\,$. Use the substitution $u=t^2\,$.

$\lim_{x\rightarrow \infty} \int_1^{x} \frac{dt}{t} = L = \lim_{x\rightarrow \infty} \int_1^{x^2} \frac{du}{2\,u} = \frac{L}{2}$

Since $L\,$ cannot be $0\,$, $L=\infty\,\qquad \square$

Theorem II

$\lim_{n\rightarrow \infty} \int_1^{\left(1+\frac{1}{n}\right)^n} \frac{dt}{t} = 1$

Let $t=u^n\,$

$\lim_{n\rightarrow \infty} \int_1^{\left(1+\frac{1}{n}\right)^n} \frac{dt}{t} = \lim_{n\rightarrow \infty}n\, \int_1^{\left(1+\frac{1}{n}\right)} \frac{du}{u}$

Since $\frac{1}{u}\,$ is monotonically decreasing, it maximum occurs at 1 and its minimum occurs at $u=\left(1+\frac{1}{n}\right)$. We can use these and the difference of the limits, $\frac{1}{n}$, to bound the integral.

$\frac{n}{n}\frac{1}{\left(1+\frac{1}{n}\right)}\le n\,\int_1^{\left(1+\frac{1}{n}\right)} \frac{du}{u} \le \frac{n}{n}$
$\lim_{n\rightarrow \infty} \int_1^{\left(1+\frac{1}{n}\right)^n} \frac{dt}{t} = \lim_{n\rightarrow \infty}n\, \int_1^{\left(1+\frac{1}{n}\right)} \frac{du}{u}=1$

Using the squeeze theorem (http://en.wikipedia.org/wiki/Squeeze_theorem). Then from the Lemma IIL $\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n,\quad n\in\mathbb{N}$ is finite since the limit of the integral is infinite for an infinite upper limit.

$\square$