Problem 34 - Essential Calculus Early Transcendentals

Question

\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=2}^{\infty}\left(\frac{-2 n}{n+1}\right)^{5 n} $$

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Understand the Series

First, we recognize the series is given by \( \sum_{n=2}^{\infty}\left(\frac{-2 n}{n+1}\right)^{5 n} \). This indicates that each term of the series is \( a_n = \left(\frac{-2 n}{n+1}\right)^{5 n} \).

Step 2: Simplify a_n

Rewrite \( a_n \) as \( \left(-1\right)^{5n} \left(\frac{2n}{n+1}\right)^{5n} = (-1)^{5n} \left(\frac{2^5 n^5}{(n+1)^5}\right)^n\). Since \((-1)^{5n} = (-1)\), this simplifies to \(\left(-1 \cdot \frac{32n^5}{(n+1)^5}\right)^n\).

Step 3: Test for Absolute Convergence

Find the absolute value of \( a_n \), which is \(\left(\frac{32n^5}{(n+1)^5}\right)^n\). Check whether the series \( \sum_{n=2}^{\infty} \left(\frac{32n^5}{(n+1)^5}\right)^n \) converges.

Step 4: Compute the Limit of the Terms

Calculate \( \lim_{n \to \infty} \left| \left(\frac{32n^5}{(n+1)^5}\right)^n \right|^{1/n} = \lim_{n \to \infty} \frac{32n^5}{(n+1)^5} = 32 \). The limit is not less than 1, indicating divergence.

Step 5: Conclusion on Convergence Tests

Since the absolute value test shows that the series diverges, the original series cannot be absolutely convergent or conditionally convergent. Therefore, the series is divergent.

Key Concepts

Absolute Convergence

When examining a series for convergence, one important aspect to check is whether it is absolutely convergent. A series \(\sum a_n\) is absolutely convergent if the series \(\sum |a_n|\), the series of its absolute values, converges. This is a stronger form of convergence because if a series is absolutely convergent, it is also convergent in the traditional sense. For example, take the series \(\sum (-1)^n \frac{1}{n!}\). To check if it's absolutely convergent, we consider \(\sum \left|(-1)^n \frac{1}{n!}\right| = \sum \frac{1}{n!}\). Since the factorial \(n!\) grows very fast, the series \(\sum \frac{1}{n!}\) is known to converge.In the context of our exercise, we attempted to find if the series \(\sum_{n=2}^{\infty} \left| \left(\frac{32n^5}{(n+1)^5} \right)^n \right|\) converges. The computation showed the limit is not less than 1, which implies divergence rather than absolute convergence.

Conditional Convergence

Conditional convergence is an interesting phenomenon. A series is conditionally convergent if it converges but does not converge absolutely. Simply put, when the series \(\sum a_n\) converges, but the series \(\sum |a_n|\) does not, the series exhibits conditional convergence.A classic example of a conditionally convergent series is the alternating harmonic series: \(\sum (-1)^{n+1} \frac{1}{n}\). This series converges by the alternating series test. However, the harmonic series \(\sum \frac{1}{n}\) diverges, meaning it does not converge absolutely.For the exercise in question, the series \(\sum_{n=2}^{\infty}\left(\frac{-2 n}{n+1}\right)^{5 n}\) was first checked for absolute convergence. When found to diverge, the next step would typically be to check traditional or conditional convergence. However, since the absolute value already shows divergence, it confirms the series is neither absolutely nor conditionally convergent.

Divergence Tests

Understanding divergence is essential when analyzing the convergence of a series. A series \(\sum a_n\) is divergent if it does not meet the criteria for convergence.Several tests can identify divergence:

• **The Divergence Test:** If \(\lim_{n \to \infty} a_n eq 0\), the series diverges. Although if \(\lim_{n \to \infty} a_n = 0\), it does not guarantee convergence.
• **Comparison Test for Divergence:** If a series \(\sum a_n\) compares unfavorably to a known divergent series, it also diverges.
• **Limit Comparison Test:** Similar to the comparison test, but more flexible, it uses the limit of the ratio \(\frac{a_n}{b_n}\) to test for divergence against other series.

In our specific case from the exercise, when applying the test for absolute convergence and finding \(\lim_{n \to \infty} \left(\frac{32n^5}{(n+1)^5}\right) = 32\), which is not less than 1, we determined that the series diverges. This finding effectively closed the discussion on potential convergence for this series.