Problem 29 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n!} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Identify the Series

We are given the series \( \sum_{n = 1}^{\infty} \frac{1}{n!} \). This is an infinite series, where the term \( a_n = \frac{1}{n!} \). Our task is to determine whether this series converges or diverges.

Step 2: Recognize the Series Type

The series \( \sum_{n = 1}^{\infty} \frac{1}{n!} \) is reminiscent of the Taylor series expansion for \( e^x \), which is \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) when \( x = 1 \). Hence, this suggests that the given series is similar to this well-known convergent series when \( x = 1 \).

Step 3: Apply the Ratio Test

The ratio test is useful for factorial expressions. Calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1}{(n+1)!} \cdot \frac{n!}{1} \right| = \lim_{n \to \infty} \frac{1}{n+1} \).

Step 4: Evaluate the Limit

The limit \( L = \lim_{n \to \infty} \frac{1}{n+1} = 0 \). Because \( L < 1 \), the ratio test indicates that the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges absolutely.

Key Concepts

Infinite Series

An infinite series is a sum of an endless sequence of terms. In mathematics, this means adding up numbers indefinitely. The question in "infinite series" is whether this sum tends to a particular number or grows without limit, which is known as convergence and divergence respectively.

• Convergence: When all terms are summed, the series approaches a specific value.
• Divergence: When summed, the series does not settle at a particular number.

In our exercise, we are dealing with the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \). To see its behavior, we need to determine if the addition of these terms will stabilize or keep increasing without bound.
Using tests and known series comparisons, we judge if this series converges or diverges. The infinite nature of the series requires these analytical tools to deduce its behavior effectively.

Factorial Expressions

Factorial expressions involve the factorial function, denoted by \( n! \). This is the product of all positive integers less than or equal to \( n \). Factorials grow very fast as \( n \) increases, which can hugely influence the behavior of the terms in a series.

• For example: \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
• The rapid increase of the denominator in expressions like \( \frac{1}{n!} \) often leads to smaller terms, which helps in convergence.

In our series \( \sum_{n=1}^{\infty} \frac{1}{n!} \), each term \( \frac{1}{n!} \) becomes very small very quickly. This is why factorial expressions are key when analyzing series such as this.

Ratio Test

The ratio test is a handy tool when dealing with series, particularly those involving factorials. It helps determine whether a series converges or diverges. The test considers the limit of the absolute value of the ratio of consecutive terms in the series.To apply the ratio test:

• Calculate \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For our series \( \sum_{n=1}^{\infty} \frac{1}{n!} \), when applying the test, the limit we obtained was \( L = 0 \). Since \( 0 < 1 \), this indicates convergence.
Using the ratio test simplifies the process of analyzing such complex series greatly.

Taylor Series

Taylor series relate to the expression of functions as infinite sums of terms. These terms are calculated from the function's derivatives at a specific point. This series expansion is incredibly useful in approximating functions with a polynomials series.The general form of a Taylor series centered at 0 (also known as a Maclaurin series) for a function \( f(x) \) is:display math \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]For example, the exponential function \( e^x \) can be expressed as:\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]When \( x = 1 \), the series becomes \( \sum_{n=0}^{\infty} \frac{1}{n!} \), which is known to converge to \( e \).
This illustrates how the given series in the exercise is similar to the Taylor series when \( x = 1 \), confirming its convergence.