sigma (infintiy and n=1) (n^2(-10)^n)/(10^nn^(8/3))

To evaluate the series 

\[
\sum_{n=1}^{\infty} \frac{n^2 (-10)^n}{10^n n^{8/3}},
\]

we can simplify and analyze the terms to understand the convergence of the series.

1. **Simplify the Series Expression:**  
   Rewrite the term as:

   \[
   \frac{n^2 (-10)^n}{10^n n^{8/3}} = \frac{n^2}{n^{8/3}} \cdot \frac{(-10)^n}{10^n}.
   \]

2. **Separate Components:**  
   Let's focus on the first component, \(\frac{n^2}{n^{8/3}}\):
   
   \[
   \frac{n^2}{n^{8/3}} = n^{2 - \frac{8}{3}} = n^{-\frac{2}{3}}.
   \]

   So the term simplifies to:

   \[
   \frac{(-10)^n}{10^n} \cdot n^{-\frac{2}{3}}.
   \]

3. **Simplify the Exponential Part:**  
   Since \(\frac{(-10)^n}{10^n} = (-1)^n\), the term can now be written as:

   \[
   (-1)^n \cdot n^{-\frac{2}{3}}.
   \]

4. **Resulting Series Form:**  
   The series becomes:

   \[
   \sum_{n=1}^{\infty} (-1)^n \cdot n^{-\frac{2}{3}}.
   \]

5. **Determine Convergence:**  
   This is an alternating series of the form \(\sum (-1)^n a_n\), where \(a_n = n^{-\frac{2}{3}}\). The convergence can be checked using the **Alternating Series Test**:
   - \(a_n = n^{-\frac{2}{3}}\) is positive for all \(n\).
   - \(a_n\) is decreasing and approaches zero as \(n \to \infty\).

   By the Alternating Series Test, this series converges.

### Conclusion:
The series \(\sum_{n=1}^{\infty} \frac{n^2 (-10)^n}{10^n n^{8/3}}\) converges.

Would you like more details on the convergence tests or additional steps?

---

1. How does the Alternating Series Test work in more depth?
2. Can we use another test to confirm convergence?
3. What happens if the exponent in \(n^{8/3}\) is changed?
4. What if the base of the exponent were different from 10?
5. How does changing the series to a non-alternating form affect convergence?

**Tip:** When analyzing series with alternating terms, always consider the Alternating Series Test to check for convergence quickly.

sigma (infinity and n=1) (n^2(-10)^n)/(10^n n^(8/3))

Explore the convergence of the infinite series \(\sum_{n=1}^{\infty} \frac{n^2 (-10)^n}{10^n n^{8/3}}\) using the Alternating Series Test. This detailed solution covers key concepts in series convergence, including simplifying terms and verifying conditions for convergence. Ideal for college-level calculus students.

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