Question

Use a calculator to compute the values of $$ 2+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} $$ for \(n=4,6,8,\) and \(10 .\) Compare each result with \(e .\)

Short answer

For n = 4, 6, 8, and 10, the sums are 2.7084, 2.7181, 2.7183, and 2.7183, respectively. Each result is approximately equal to \( e \).

Step-by-step solution

Introduction

We need to compute the sum of the series \[ 2 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} \]for values of \( n = 4, 6, 8, 10 \), and compare each result with the value of the mathematical constant \( e \). The value of \( e \) is approximately 2.71828.

- Compute the sum for \( n = 4 \)

For \( n = 4 \), the series sum is \[ 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \]Calculate each term:\( 2 = 2 \)\( \frac{1}{2!} = \frac{1}{2} = 0.5 \)\( \frac{1}{3!} = \frac{1}{6} \approx 0.1667 \)\( \frac{1}{4!} = \frac{1}{24} \approx 0.0417 \)Adding these values gives \[ 2 + 0.5 + 0.1667 + 0.0417 = 2.7084 \]Compare with \( e \): \[ 2.7084 \approx 2.71828 \]

- Compute the sum for \( n = 6 \)

For \( n = 6 \), the series sum is \[ 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} \]Calculate each additional term:\( \frac{1}{5!} = \frac{1}{120} \approx 0.0083 \)\( \frac{1}{6!} = \frac{1}{720} \approx 0.0014 \)Adding these to the previous sum gives \[ 2.7084 + 0.0083 + 0.0014 = 2.7181 \]Compare with \( e \): \[ 2.7181 \approx 2.71828 \]

- Compute the sum for \( n = 8 \)

For \( n = 8 \), the series sum is \[ 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + \frac{1}{8!} \]Compute the additional terms:\( \frac{1}{7!} = \frac{1}{5040} \approx 0.0001984 \)\( \frac{1}{8!} = \frac{1}{40320} \approx 0.0000248 \)Adding these to the previous sum gives \[ 2.7181 + 0.0001984 + 0.0000248 = 2.7183 \]Compare with \( e \): \[ 2.7183 \approx 2.71828 \]

- Compute the sum for \( n = 10 \)

For \( n = 10 \), the series sum is \[ 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + \frac{1}{8!} + \frac{1}{9!} + \frac{1}{10!} \]Compute the additional terms:\( \frac{1}{9!} = \frac{1}{362880} \approx 0.0000027 \)\( \frac{1}{10!} = \frac{1}{3628800} \approx 0.0000003 \)Adding these to the previous sum gives \[ 2.7183 + 0.0000027 + 0.0000003 = 2.7183 \]Compare with \( e \): \[ 2.7183 \approx 2.71828 \]

Key concepts

Factorials

Factorials are a fundamental concept in mathematics, especially in combinatorics and series. The factorial of a non-negative integer is the product of all positive integers less than or equal to that integer. It is denoted by an exclamation mark (!). For example, 3! = 3 × 2 × 1 = 6.
They're crucial in the given series because each term involves a factorial in the denominator. Factorials grow very quickly, which means higher terms in the series contribute less, helping the series converge.

Exponential Constant

The exponential constant, often denoted as \( e \), is approximately equal to 2.71828. It's a fundamental constant in mathematics, particularly in calculus and complex analysis. You encounter \( e \) in many natural processes, including compound interest, population growth, and decay.
Mathematically, \( e \) can be defined as the sum of the series: \[ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + \frac{1}{8!} + \frac{1}{9!} + ... \]
This series converges to the value of \( e \) as more terms are added.

Convergence of Series

When dealing with series, convergence is an important concept. A series converges if its sum approaches a finite value as more terms are added. In our exercise, the series '\[ 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... \]' converges to the value of \( e \).
This is because the additional terms (\( \frac{1}{n!} \)) become extremely small as \( n \) increases. For instance, \( \frac{1}{10!} \) is only about 0.0000003. Due to this property, the series sum gets very close to \( e \) even with only a few terms.

Approximation of e

The exercise of summing the series for different values of \( n \) (such as 4, 6, 8, and 10) shows how quickly the series approaches \( e \). With \( n = 4 \), the sum is already close to \( e \) at 2.7084. When \( n = 6 \), the sum is 2.7181. By the time \( n = 10 \), the sum is virtually indistinguishable from \( e \) at 2.7183.
This demonstrates the efficiency of the series in approximating \( e \). It also highlights how each new term has less impact, speeding up the convergence to the actual value of \( e \).