Question

In the following exercises, use an appropriate test to determine whether the series converges. $$ \sum_{n=1}^{\infty} \frac{(n-1)^{n}}{(n+1)^{n}} $$

Short answer

The series diverges by approaching form suggestions and ratio analysis.

Step-by-step solution

Analyze the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{(n-1)^{n}}{(n+1)^{n}} \). First, consider the general term \( a_n = \frac{(n-1)^{n}}{(n+1)^{n}} \). Notice that both the numerator and the denominator are exponential terms.

Consider the Ratio Test

We'll use the Ratio Test to determine the convergence. This test states that for a series \( \sum a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \) and \( L < 1 \), the series converges. If \( L > 1 \), the series diverges.

Compute the Ratio

Calculate \( \frac{a_{n+1}}{a_n} \) where \( a_{n+1} = \frac{n^{n+1}}{(n+2)^{n+1}} \). We have:\[\frac{a_{n+1}}{a_n} = \frac{\left(\frac{n^{n+1}}{(n+2)^{n+1}}\right)}{\left(\frac{(n-1)^n}{(n+1)^n}\right)} = \frac{n^{n+1} \cdot (n+1)^n}{(n-2)^n (n+2)^{n+1}}\]

Simplify the Ratio Expression

Break down the expression:\[\frac{n^{n+1} (n+1)^n}{(n-1)^n (n+2)^{n+1}} = \left(\frac{n}{n+2}\right)^{n+1} \cdot \left(\frac{n+1}{n-1}\right)^n\]

Apply the Limit

Now take the limit as \( n \to \infty \):\[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n}{n+2}\right)^{n+1} \cdot \left(\frac{n+1}{n-1}\right)^n\]Using the fact that \(\frac{n}{n+2} \approx 1 \) and \(\frac{n+1}{n-1} \approx 1\), analyze the behavior as \( n \to \infty \).

Evaluate the Exponential Terms

The terms \( \left(\frac{n}{n+2}\right)^{n+1} \) and \( \left(\frac{n+1}{n-1}\right)^n \) both tend towards 1 as \( n \to \infty \). Thus, the limit of the ratio is 1.

Final Step: Conclude with the Divergence

Since \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1 \), the Ratio Test is inconclusive. However, observe that the form \( \left(\frac{n-1}{n+1}\right)^n \) suggests divergence due to the expression tending to a fraction less than 1 raised to increasing powers, which pushes the terms towards 0 slower than any converging series (intuitively similar to divergent geometric series with a fractional base). More sophisticated tests might typically be used to further analyze such forms.

Key concepts

Ratio Test

The Ratio Test is a common method used to test the convergence of a series. It helps to identify whether the terms of a series get sufficiently small as the series progresses. For a series \( \sum a_n \), the test considers the limit of the absolute value of the ratio of successive terms:

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

Understanding the Ratio Test is vital since it helps determine how the sequence of numbers in the series approaches zero. If they approach zero quickly enough (\( L < 1 \)), the series will converge. In this problem, however, the calculation showed \( L = 1 \) and thus, the test did not provide a clear conclusion on its own.

Divergence

Divergence in series occurs when the terms do not approach zero or sum up to a finite value as we add more and more of them. Even if the Ratio Test is inconclusive, as seen when \( L = 1 \), divergence may still be detected through other means.For the series \( \sum_{n=1}^{\infty} \frac{(n-1)^{n}}{(n+1)^{n}} \), we observed that the series has a particular form that diverges. This is similar to a geometric series but not converging, because the factorial growth in the exponents results in a pattern where terms do not diminish outrightly to a negligible value. Many series that have a similar pattern \( \left( \frac{n-1}{n+1} \right)^n \) suggest divergence as this expression tends towards a fraction less than 1 raised to increasingly larger powers.Understanding when a series diverges can sometimes involve recognizing patterns or using more advanced tests beyond the Ratio Test, especially if terms grow or decline slower than expected for convergence.

Exponential Functions

Exponential functions appear when dealing with series involving terms raised to powers, especially in rapidly growing sequences. In this exercise, both the numerator \( (n-1)^{n} \) and the denominator \( (n+1)^{n} \) include exponential expressions.Key characteristics occur because:

• As \( n \) increases, an expression like \( (n-1)^{n} \) grows very quickly.
• Considering a geometric form with bases close to one (as series progresses), even slight deviations in the base can lead to different results of convergence or divergence.

In this problem's context, analyzing such exponential relationships helps one understand series behavior better and ultimately make informed conclusions about convergence tendencies.These functions underline how small changes in the base values impact the growth of the entire term. Hence, exponential expressions in series are a crucial aspect of convergence analysis.