Math Problem Statement

Prove that the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{8^n} \) is convergent.

Solution

To prove that the series

n=1(1)n+1n8n\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{8^n}

is convergent, we can again examine this series using the Alternating Series Test and consider absolute convergence since the term n8n\frac{n}{8^n} suggests an exponential decay that often leads to absolute convergence.

Step 1: Absolute Convergence Check

To determine absolute convergence, consider the absolute value of the terms in the series:

n=1(1)n+1n8n=n=1n8n.\sum_{n=1}^{\infty} \left| \frac{(-1)^{n+1} n}{8^n} \right| = \sum_{n=1}^{\infty} \frac{n}{8^n}.

This is now a positive series, which we can test for convergence using the Ratio Test or Root Test, both of which are effective for series involving exponential terms.

Applying the Ratio Test

The Ratio Test states that a series an\sum a_n converges absolutely if

limnan+1an<1.\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.

Here, let an=n8na_n = \frac{n}{8^n}. Then,

an+1=n+18n+1.a_{n+1} = \frac{n+1}{8^{n+1}}.

So,

an+1an=n+18n+1n8n=n+18n.\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{8^{n+1}}}{\frac{n}{8^n}} = \frac{n+1}{8n}.

Taking the limit as nn \to \infty:

limnn+18n=18<1.\lim_{n \to \infty} \frac{n+1}{8n} = \frac{1}{8} < 1.

Since this limit is less than 1, the Ratio Test confirms that the series n=1n8n\sum_{n=1}^{\infty} \frac{n}{8^n} converges absolutely.

Conclusion

Since the series converges absolutely, the original series

n=1(1)n+1n8n\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{8^n}

also converges.

Would you like further details or have any specific questions?

Here are five related questions for deeper exploration:

  1. What other convergence tests can be applied to this series for verification?
  2. How does the Ratio Test differ from the Root Test, especially in terms of application to exponential terms?
  3. Why does absolute convergence imply convergence of the original series?
  4. Can we evaluate the exact sum of this series, or is it only possible to confirm convergence?
  5. What role does the exponential decay play in the convergence of the series?

Tip: Exponential terms in a series often allow for straightforward convergence testing using the Ratio or Root Test due to the rapid decay they introduce.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Absolute Convergence
Alternating Series Test
Ratio Test

Formulas

Ratio Test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \)
Series representation: \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n}{8^n} \)

Theorems

Alternating Series Test
Ratio Test

Suitable Grade Level

Undergraduate Calculus or Advanced Placement (AP) Calculus