Question

Determine the convergence of the given series. State the test used; more than one test may be appropriate. $$\sum_{n=1}^{\infty} \frac{\ln n}{n !}$$

Short answer

The series converges by the ratio test.

Step-by-step solution

Analyze the Form of the Series

The series given is \( \sum_{n=1}^{\infty} \frac{\ln n}{n!} \). This is a series with a factorial in the denominator. When considering series with factorials, the ratio test is often useful.

Apply the Ratio Test

To apply the ratio test, calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| \), where \( a_n = \frac{\ln n}{n!} \). Compute:\[\frac{a_{n+1}}{a_{n}} = \frac{\frac{\ln(n+1)}{(n+1)!}}{\frac{\ln n}{n!}} = \frac{\ln(n+1)}{n+1} \cdot \frac{1}{\ln n} = \frac{\ln(n+1)}{(n+1) \cdot \ln n}\]Now, compute the limit:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| = \lim_{n \to \infty} \frac{\ln(n+1)}{(n+1) \cdot \ln n}\]

Evaluate the Limit

Evaluate the limit:\[\lim_{n \to \infty} \frac{\ln(n+1)}{(n+1) \cdot \ln n} \]As \( n \to \infty \), the terms \( \ln(n+1) \) and \( \ln n \) are approximately equal, but \( n+1 \to \infty \) dominates the numerator. This limit approaches zero.

Conclude with the Ratio Test

Since \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1 \), the series \( \sum_{n=1}^{\infty} \frac{\ln n}{n!} \) converges by the ratio test.

Key concepts

Ratio Test

The ratio test is a powerful tool to determine the convergence or divergence of an infinite series. It involves examining the limit of the absolute value of the ratio between consecutive terms of the series.

• Begin with the series \ \( \sum_{n=1}^{\infty} a_n \ \).
• The ratio test involves computing the limit \ \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \ \).
• If this limit is less than 1, the series converges absolutely.
• If the limit is greater than 1 or infinite, the series diverges.
• If the limit equals 1, the test is inconclusive and another method is needed.

This test is especially helpful when the terms of the series involve factorials or exponential functions, which grow or shrink quickly. In the given exercise, we applied the ratio test to the series \ \( \sum_{n=1}^{\infty} \frac{\ln n}{n!} \ \) and found that the limit of the ratio is zero, thereby deducing that the series converges.

Factorials in Series

Factorials often appear in series and sequences, particularly those involving combinatorials and growth rates. The factorial of a non-negative integer \( n \) is the product of all positive integers less than or equal to \( n \), denoted as \( n! \).

• The factorial function increases very rapidly as \( n \) grows larger.
• This rapid growth often means series involving factorials in the denominator, like \ \( \sum_{n=1}^{\infty} \frac{\ln n}{n!} \ \), will converge.
• The reason is that the factorial grows faster than exponential functions or logarithmic functions used in the numerator.

Understanding the behavior of factorials helps simplify the application of tests like the ratio test, as seen in the exercise, where factorials effectively lead the ratio to zero, indicating convergence.

Infinite Series

An infinite series is the sum of an infinite sequence of terms. In mathematical notation, an infinite series might look like \ \( \sum_{n=1}^{\infty} a_n \ \).

• One major goal when dealing with infinite series is determining whether they converge or diverge.
• A series converges if the sum approaches a finite number as more terms are added.
• Otherwise, the series is said to diverge, meaning it grows without bound or oscillates indefinitely.
• Various tests, like the integral test, comparison test, and ratio test, can be applied to decide the convergence of different types of series.

The given exercise involves such a series that incorporates logarithmic and factorial components. After applying the ratio test to the series \ \( \sum_{n=1}^{\infty} \frac{\ln n}{n!} \ \), it was determined to converge, demonstrating how infinite series can be analyzed using strategic testing methods.