https://file.aiquickdraw.com/mathbot/file/1730860479280ihdmqeln.jpg

To prove that the series 

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

is convergent, we can again examine this series using the **Alternating Series Test** and consider **absolute convergence** since the term \( \frac{n}{8^n} \) 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:

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

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 \( \sum a_n \) converges absolutely if 

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

Here, let \( a_n = \frac{n}{8^n} \). Then,

\[
a_{n+1} = \frac{n+1}{8^{n+1}}.
\]

So,

\[
\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{8^{n+1}}}{\frac{n}{8^n}} = \frac{n+1}{8n}.
\]

Taking the limit as \( n \to \infty \):

\[
\lim_{n \to \infty} \frac{n+1}{8n} = \frac{1}{8} < 1.
\]

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

### Conclusion
Since the series converges absolutely, the original series 

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

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.

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

Learn how to prove the convergence of the series \( \sum \frac{(-1)^{n+1} n}{8^n} \). This solution explores the concepts of absolute convergence, the Ratio Test, and the Alternating Series Test, providing a detailed step-by-step proof suitable for advanced calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation