Q. 7 - Calculus

Question

Use a calculator to find the sum of the first six terms of the sum from the previous problem, and compare this sum with your calculator’s best approximation of the number e.

Step-by-Step Solution

Answer

The value of the first six-term of the expanded form of expression ${\left(1+\frac{1}{n}\right)}^n$with a limit $n\to \infty$is the same as the value of e up to two decimal places.

Step 1. Given information.

The given expanded form of expression as $n\to \infty$is ${\left(1+\frac{1}{n}\right)}^n=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots .$

Step 2. Value of sum of the first six terms.

The sum of the first six-term of the expanded form of expression${\left(1+\frac{1}{n}\right)}^n$ with a limit $n\to \infty$is as follows.

$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}=2.176$

value of e is $2.718.$

So the value of the first six-term of the expanded form of expression ${\left(1+\frac{1}{n}\right)}^n$with a limit $n\to \infty$is the same as the value of e up to two decimal places.