Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{n^{2}}{n !}\)

Step-by-Step Solution

Answer

The sum is \( 2e \).

Step 1: Understand the Expression

The given expression is a series: \( \sum_{n=1}^{\infty} \frac{n^2}{n !} \). We need to find the sum of this series if it converges. The notation \( n! \) represents the factorial of \( n \), which is the product of all positive integers up to \( n \).

Step 2: Test for Convergence

To determine if the series converges, we can use the ratio test. Let's find the limit of the absolute value of the ratio of successive terms: \[ \lim_{n \to \infty} \left| \frac{(n+1)^2}{(n+1)!} \cdot \frac{n!}{n^2} \right| = \lim_{n \to \infty} \frac{(n+1)^2}{n^2(n+1)} = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^3 + n^2} \] Simplify:\[ = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{n + 1} = \lim_{n \to \infty} \frac{1}{n+1} = 0 \] Since this limit is 0, which is less than 1, the ratio test confirms the series converges.

Step 3: Find the Sum using Known Series

The series resembles the exponential function because \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \). We can differentiate this known series representation to help us find the sum.Starting with \( e^x \), differentiate with respect to \( x \): \[ \frac{d}{dx}(e^x) = e^x \Rightarrow \frac{d}{dx}\left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) = \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n!} \] Multiply by \( x \): \[ \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n!} \cdot x = \sum_{n=1}^{\infty} \frac{nx^n}{n!} = xe^x \] Differentiate again: \[ \frac{d}{dx}(xe^x) = e^x + xe^x \Rightarrow \sum_{n=1}^{\infty} \frac{n^2x^{n-1}}{n!} \] Multiply by \( x \): \[ \sum_{n=1}^{\infty} \frac{n^2x^n}{n!} = (1+x)e^x \] Set \( x = 1 \) to solve the original sum:\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = (1+1)e^1 = 2e \]

Step 4: Conclude the Sum

Based on our calculations using the series properties of the exponential function, the sum of the series \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \) is \( 2e \).

Key Concepts

Understanding the Ratio Test

The ratio test is a handy tool used to determine whether a series converges or diverges. To use it, you calculate the limit of the absolute value of the ratio of consecutive terms in the series.
Here's how it works:

• Consider a series \( \sum_{n=1}^{\infty} a_n \).
• Calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

If \( L < 1 \), the series converges absolutely. If \( L > 1 \) or \( L = \infty \), the series diverges. If \( L = 1 \), the test is inconclusive.
This test is particularly useful for series with factorials, as factorials grow rapidly, often making terms smaller quickly.

The Concept of Convergence

Convergence refers to the idea that as we add more terms of the series, we get closer and closer to a specific value. If a series converges, it means there is a finite sum you can approach by adding enough terms.
When using the ratio test, as in our example, it's crucial to simplify expressions to clearly see if the series converges.
In the series \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \), the ratio test showed that the limit approached \( 0 \), which is less than \( 1 \), confirming convergence.

• This means that the sum of the series is finite.
• We begin with an infinite number of terms but manage to find this total sum.

Factorials in Infinite Series

Factorials, denoted by \( n! \), are instrumental in understanding series like \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \). A factorial is the product of all positive integers up to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).
Why do factorials matter in series?

• They grow faster than exponential terms, usually making series converge faster.
• They help simplify calculations by forming neat, cancelable structures when using the ratio test.

In our original series, the factorial in the denominator causes rapid term reduction, aiding convergence as demonstrated through the ratio test. This implies that factorials act as a powerful balancing force when evaluating infinite series.

Exploring the Exponential Function

The exponential function, represented by \( e^x \), is a foundational mathematical function with properties that connect deeply with infinite series.
In our solution, we recognized the resemblance of the series to portions of the exponential function.
The exponential function's Taylor series expansion:

• Looks like this: \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
• These functions enable conversion of complex series into familiar forms.

By differentiating and manipulating this expansion, we find that the sum of our original series \( \sum_{n=1}^{\infty} \frac{n^2}{n!} \) equals \( (1+1)e \) or simply \( 2e \).
This showcases the remarkable power of exponential functions in simplifying the evaluation of certain infinite series.