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?

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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.