Question

Determine the convergence of the given series. State the test used; more than one test may be appropriate. $$\sum_{n=1}^{\infty} \frac{n^{3}}{(\ln n)^{n}}$$

Short answer

The series converges by the Ratio Test.

Step-by-step solution

Recognize the form of the series

Examine the given series: \( \sum_{n=1}^{\infty} \frac{n^{3}}{(\ln n)^n} \). Notice that it involves both a polynomial in the numerator and a logarithmic function raised to the power of \( n \) in the denominator.

Determine the suitability of the Ratio Test

For series like this, the Ratio Test can be very effective. The Ratio Test involves examining the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as \( n \to \infty \), where \( a_n = \frac{n^3}{(\ln n)^n} \).

Calculate the Ratio of Consecutive Terms

Calculate \( a_{n+1} = \frac{(n+1)^3}{(\ln(n+1))^{n+1}} \) and find \( \left| \frac{a_{n+1}}{a_n} \right| \) which simplifies to \( \frac{(n+1)^3}{n^3} \cdot \frac{(\ln n)^n}{(\ln(n+1))^{n+1}} \).

Simplify the Expression for the Ratio

Begin by simplifying \( \frac{(n+1)^3}{n^3} \approx 1 + \frac{3}{n} \) for large \( n \). Then explore \( \frac{(\ln n)^n}{(\ln(n+1))^{n+1}} \approx (\frac{\ln n}{\ln(n+1)})^n \cdot \frac{1}{\ln(n+1)} \).

Find the Limit of the Ratio

As \( n \to \infty \), \( \frac{\ln n}{\ln(n+1)} \to 1 \), making the exponential \( (\frac{\ln n}{\ln(n+1)})^n \to e^{-1} \), and \( \frac{1}{\ln(n+1)} \approx \frac{1}{\ln n} \). The entire expression \( \left| \frac{a_{n+1}}{a_n} \right| \to 0 \).

Applying the Ratio Test Conclusion

Since the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \) exists and is less than 1, the series \( \sum_{n=1}^{\infty} \frac{n^3}{(\ln n)^n} \) converges absolutely by the Ratio Test.

Key concepts

Ratio Test

The Ratio Test is a powerful method used to determine whether an infinite series converges or not. This test involves taking the limit of the absolute value of the ratio of successive terms in the series. If you have a series \( \sum a_n \), you calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• If the result is less than 1, the series converges absolutely.
• If it is greater than 1 or infinity, the series diverges.
• If the result is exactly 1, the test is inconclusive.

This test is particularly effective for series where each term is a product of factorial, exponentials, or powers. By reducing the problem to simpler calculations and comparisons, the Ratio Test provides a clear indication of convergence or divergence.

Convergent Series

A convergent series is an infinite series whose terms approach a limited sum as more and more terms are added. In other words, given a series \( \sum_{n=1}^{\infty} a_n \), it is convergent if the sum \( S = a_1 + a_2 + a_3 + \dots \) approaches a finite limit as the number of terms goes to infinity.
Convergence is an important property, as it means the series effectively settles down to a stable value, unlike divergent series which escape to infinity. For example, convergent series are key in representing functions or in approximating values in mathematical analysis.
Different tests, like the Ratio Test, the Root Test, or the Integral Test, can be used to determine the convergence of a series. Understanding when and how to use these tests can critically enhance one's ability in resolving series problems.

Infinite Series

An infinite series is a sum of infinitely many terms, written in the form \( S = a_1 + a_2 + a_3 + \ldots \). Such series are fundamental in mathematics, particularly in calculus and analysis.
Many phenomena in physics and engineering rely on infinite series to model complex systems and behaviors.
Examples include:

• Geometric series, such as \( \sum_{n=0}^{\infty} ar^n \), where \( |r| < 1 \) ensures convergence.
• The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), known for its divergence despite each term decreasing.

Understanding the behavior of infinite series, and knowing when they converge or diverge, is crucial for applied sciences and theoretical mathematics alike.