#  lim_(n->oo)(1+1/n)^n =  ?

Sep 22, 2016

Let ${a}_{n}$ and ${b}_{n}$, $n \setminus \in N$, are defined as:

${a}_{n} = {\left(1 + \frac{1}{n}\right)}^{n}$,

${b}_{n} = {\left(1 + \frac{1}{n}\right)}^{n + 1} = {a}_{n} \left(1 + \frac{1}{n}\right) .$

It is obvious that:

${\lim}_{n \to \infty} {a}_{n} = \frac{{\lim}_{n \to \infty} {b}_{n}}{{\lim}_{n \to \infty} \left(1 + \frac{1}{n}\right)} = {\lim}_{n \to \infty} {b}_{n} .$

We'll show that sequence ${b}_{n}$ is decreasing.

${b}_{n - 1} / {b}_{n} = {\left(1 + \frac{1}{n - 1}\right)}^{n} / {\left(1 + \frac{1}{n}\right)}^{n + 1} = {\left(\frac{n}{n - 1}\right)}^{n} / {\left(\frac{n + 1}{n}\right)}^{n + 1} = {n}^{2 n + 1} / \left({\left(n - 1\right)}^{n} {\left(n + 1\right)}^{n + 1}\right)$

${b}_{n - 1} / {b}_{n} = {\left({n}^{2} / \left({n}^{2} - 1\right)\right)}^{n} \cdot \frac{n}{n + 1} = {\left(1 + \frac{1}{{n}^{2} - 1}\right)}^{n} \cdot \frac{n}{n + 1}$

Using Bernoulli's inequality ${\left(1 + x\right)}^{n} > 1 + n x$ it follows:

${\left(1 + \frac{1}{{n}^{2} - 1}\right)}^{n} > 1 + \frac{n}{{n}^{2} - 1}$ and:

${b}_{n - 1} / {b}_{n} > \left(1 + \frac{n}{{n}^{2} - 1}\right) \cdot \frac{n}{n + 1} > \left(1 + \frac{n}{n} ^ 2\right) \cdot \frac{n}{n + 1}$

${b}_{n - 1} / {b}_{n} > \frac{n \left(n + 1\right)}{n} ^ 2 \cdot \frac{n}{n + 1} = 1$

So, the sequence ${b}_{n}$ is decreasing.

Using Bernoulli's inequality again:

${b}_{n} = {\left(1 + \frac{1}{n}\right)}^{n + 1} > 1 + \frac{n + 1}{n} = 1 + 1 + \frac{1}{n} = 2 + \frac{1}{n} > 2$

it follows that ${b}_{n}$ has a lower bound. ${b}_{n}$ is convergent.

Using inequality of arithmetic and geometric means:

$\sqrt[n + 1]{{\left(1 + \frac{1}{n}\right)}^{n} \cdot 1} < \frac{n \left(1 + \frac{1}{n}\right) + 1}{n + 1} = \frac{n + 2}{n + 1}$

it follows:

${\left(1 + \frac{1}{n}\right)}^{n} < {\left(\frac{n + 2}{n + 1}\right)}^{n + 1} = {\left(1 + \frac{1}{n + 1}\right)}^{n + 1} ,$

${a}_{n} < {a}_{n + 1}$

and ${a}_{n}$ is increasing.

${a}_{n} = {\left(1 + \frac{1}{n}\right)}^{n} < {\left(1 + \frac{1}{n}\right)}^{n + 1} = {b}_{n}$

It follows that ${a}_{n}$ has an upper bound (because ${b}_{n}$ is decreasing and has a lower bound). Sequence ${a}_{n}$ is increasing and has an upper bound so it's convergent.

Finally,

$e = {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} .$

Sep 22, 2016

Let some value $x = {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}$, where $n$ is an arbitrary variable (i.e. does not have to be an integer!). Then the fastest way to proceed is as follows.

$\ln x = \ln \left({\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}\right)$

$\ln x = {\lim}_{n \to \infty} \left(n \ln \left(1 + \frac{1}{n}\right)\right)$

$\ln x = {\lim}_{n \to \infty} \left(\ln \frac{1 + \frac{1}{n}}{\frac{1}{n}}\right)$

By inspection, we've converted the right side into a $\frac{0}{0}$ form. Therefore, we can use L'Hopital's rule.

$\ln x = {\lim}_{n \to \infty} \left(\frac{\frac{1}{1 + \frac{1}{n}} \cdot \cancel{- \frac{1}{n} ^ 2}}{\cancel{- \frac{1}{n} ^ 2}}\right)$

$\ln x = {\lim}_{n \to \infty} \left(\frac{1}{1 + \frac{1}{n}}\right)$

wherein the $\frac{1}{n}$ term vanishes as $n \to \infty$.

$\ln x = {\lim}_{n \rightarrow \infty} \left(\frac{1}{1 + 0}\right)$

$\ln x = {\lim}_{n \rightarrow \infty} 1$

$\ln x = 1$

Thus, we undo the natural logarithm to get:

$\textcolor{b l u e}{{\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}}$

$= {e}^{\ln \left({\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}\right)}$

$= {e}^{\ln x} = x = {e}^{1} = \textcolor{b l u e}{e}$.

Sep 22, 2016

${\left(1 + \frac{1}{n}\right)}^{n} < {\left(1 + \frac{1}{n + 1}\right)}^{n + 1}$ is increasing for $n \to \infty$

(1+1/n)^n = 1+1/(1!)n/n+1/(2!)(n/n)((n-1)/n) + 1/(3!)(n/n)((n-1)/n)((n-2)/n)+cdots

((n-1)/n)(1+1/n)^n le 1-1+ (n-1)/n+1/(1!)(n-1)/n+1/(2!)((n-1)/n)^2+1/(3!)((n-1)/n)^3 + 1/(4!)((n-1)/n)^4+cdots +

((n-1)/n)(1+1/n)^n le (n-1)/n-1+sum_(k=0)^n1/(k!)((n-1)/n)^k

so

lim_(n->oo)((n-1)/n)(1+1/n)^n le lim_(n->oo)(-1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)((n-1)/n)^k lelim_(n->oo)( -1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)

Finally

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

and

${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} \le e$