To determine whether the series

$\sum_{n=1}^{\infty} (-1)^n n^2 e^{-5n}$

converges or diverges, let's analyze it by breaking down its components and testing for convergence using appropriate tests.

Step 1: Examine the Terms

The terms of the series are of the form $a_n = (-1)^n n^2 e^{-5n}.$ This is an alternating series due to the factor $(-1)^n$, which means the terms switch signs as $n$ increases.

Step 2: Absolute Convergence Test

To determine convergence, we can first test for absolute convergence by examining the absolute value of the terms: $|a_n| = n^2 e^{-5n}.$ If the series of $|a_n|$ converges, then the original series will also converge absolutely.

Applying the Ratio Test

The ratio test is useful for series with factorials or exponential factors. The ratio test states that for a series $\sum a_n$, if

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L,$

then:

• If $L < 1$, the series converges absolutely.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

Let's apply the ratio test to $|a_n| = n^2 e^{-5n}$.

$\frac{|a_{n+1}|}{|a_n|} = \frac{(n+1)^2 e^{-5(n+1)}}{n^2 e^{-5n}} = \frac{(n+1)^2}{n^2} \cdot e^{-5}.$

Now, as $n \to \infty$:

1. The ratio $\frac{(n+1)^2}{n^2}$ approaches 1.
2. Thus, $\frac{|a_{n+1}|}{|a_n|} \approx e^{-5}$, a constant less than 1.

Since $e^{-5} < 1$, the ratio test confirms that $\sum |a_n|$ converges. This means the series converges absolutely.

Conclusion

Because the series converges absolutely, we conclude that the original series

$\sum_{n=1}^{\infty} (-1)^n n^2 e^{-5n}$

converges.

Would you like more details or have any questions?

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Here are five related questions to explore this concept further:

1. What is conditional convergence, and how does it differ from absolute convergence?
2. How does the alternating series test differ from the ratio test?
3. Can we apply the root test to this series? If so, how does it work?
4. How would we analyze a series with a factorial term, such as $n! e^{-n}$?
5. What is the intuition behind why exponential decay (like $e^{-5n}$) often leads to convergence?

Tip: The ratio test is often best for series with exponential or factorial terms, as it helps reveal rapid growth or decay.