Chapter 1: Q. 6 (page 151)

Show that as $n \to \infty$ we would expect the preceding expansion to approach
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots .$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

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

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

Step 2. Verification.

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

${(1 + \frac{1}{n})}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) 1^{n} {(\frac{1}{n})}^{0} + (\begin{matrix} n \\ 1 \end{matrix}) 1^{n - 1} {(\frac{1}{n})}^{1} + (\begin{matrix} n \\ 2 \end{matrix}) 1^{n - 2} {(\frac{1}{n})}^{2} + \dots + (\begin{matrix} n \\ n \end{matrix}) 1^{0} {(\frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n} = (\frac{n!}{0! (n - 0)!}) {(\frac{1}{n})}^{0} + (\frac{n!}{1! (n - 1)!}) {(\frac{1}{n})}^{1} + (\frac{n!}{2! (n - 2)!}) {(\frac{1}{n})}^{2} + \dots + (\frac{n!}{n! (n - n)!}) {(\frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n} = 1 + (\frac{n!}{1! (n - 1)!}) {(\frac{1}{n})}^{1} + (\frac{n!}{2! (n - 2)!}) {(\frac{1}{n})}^{2} + \dots + {(\frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n} = 1 + \frac{1}{1!} + \frac{1}{2!} (\frac{n - 1}{n}) + \frac{1}{3!} (\frac{(n - 1) (n - 2)}{n^{2}}) + \frac{1}{4!} (\frac{(n - 1) (n - 2) (n - 3)}{n^{3}}) + \frac{1}{5!} (\frac{(n - 1) (n - 2) (n - 3) (n - 4)}{n^{4}}) \dots$

Step 3. Taking limit.

Take limit $n \to \infty .$

$\lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = {(1 + \frac{1}{n})}^{n} = 1 + \frac{1}{1!} + \frac{1}{2!} (1) + \frac{1}{3!} (1) + \frac{1}{4!} (1) + \frac{1}{5!} (1) \dots \lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = {(1 + \frac{1}{n})}^{n} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} \dots$

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

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Show that as $n \to \infty$ we would expect the preceding expansion to approach
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots .$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Verification.

Step 3. Taking limit.

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Show that as n→∞we would expect the preceding expansion to approach1+11!+12!+13!+14!+15!+⋯.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Verification.

Step 3. Taking limit.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Show that as $n \to \infty$ we would expect the preceding expansion to approach
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots .$