Q. 6 - Calculus

Question

Show that as $n\to \infty$we would expect the preceding expansion to approach

$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots .$

Step-by-Step Solution

Answer

As $n\to \infty ,$the expression ${\left(1+\frac{1}{n}\right)}^n$approach $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\cdots .$

Step 1. Given information.

The given expression is ${\left(1+\frac{1}{n}\right)}^n.$

The given approaching expression of ${\left(1+\frac{1}{n}\right)}^n$is $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots .$

Step 2. Verification.

Expand the expression ${\left(1+\frac{1}{n}\right)}^n$according to the Binomial Theorem.

$$\begin{gathered} {\left(1+\frac{1}{n}\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)1^n{\left(\frac{1}{n}\right)}^0+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}{\left(\frac{1}{n}\right)}^1+\left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}{\left(\frac{1}{n}\right)}^2+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)1^0{\left(\frac{1}{n}\right)}^n \\ {\left(1+\frac{1}{n}\right)}^n=\left(\frac{n!}{0!(n-0)!}\right){\left(\frac{1}{n}\right)}^0+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +\left(\frac{n!}{n!(n-n)!}\right){\left(\frac{1}{n}\right)}^n \\ {\left(1+\frac{1}{n}\right)}^n=1+\left(\frac{n!}{1!(n-1)!}\right){\left(\frac{1}{n}\right)}^1+\left(\frac{n!}{2!(n-2)!}\right){\left(\frac{1}{n}\right)}^2+\cdots +{\left(\frac{1}{n}\right)}^n \\ {\left(1+\frac{1}{n}\right)}^n=1+\frac{1}{1!}+\frac{1}{2!}\left(\frac{n-1}{n}\right)+\frac{1}{3!}\left(\frac{\left(n-1\right)\left(n-2\right)}{n^2}\right)+\frac{1}{4!}\left(\frac{\left(n-1\right)\left(n-2\right)\left(n-3\right)}{n^3}\right)+\frac{1}{5!}\left(\frac{\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}{n^4}\right)\cdots \end{gathered}$$

Step 3. Taking limit.

Take limit $n\to \infty .$

$$\begin{gathered} \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n={\left(1+\frac{1}{n}\right)}^n=1+\frac{1}{1!}+\frac{1}{2!}\left(1\right)+\frac{1}{3!}\left(1\right)+\frac{1}{4!}\left(1\right)+\frac{1}{5!}\left(1\right)\cdots \\ \underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n={\left(1+\frac{1}{n}\right)}^n=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\cdots \end{gathered}$$

So as $n\to \infty ,$the expression ${\left(1+\frac{1}{n}\right)}^n$approach $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\cdots .$