The expression you're considering is $n^2 e^{-n}$, and we want to determine if the series or the function (depending on your context) converges or diverges.

Case 1: Convergence of the Series

If you're referring to the series $\sum_{n=1}^{\infty} n^2 e^{-n}$, we can apply the ratio test or comparison test to determine convergence.

Ratio Test:

For the ratio test, we compute the limit of the ratio of successive terms:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ where $a_n = n^2 e^{-n}$. We now calculate this ratio:

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

Simplifying:

$\frac{a_{n+1}}{a_n} = \left(1 + \frac{1}{n}\right)^2 e^{-1}$

As $n \to \infty$, $\left(1 + \frac{1}{n}\right)^2 \to 1$, so:

$L = e^{-1} \approx 0.3679$

Since $L < 1$, the ratio test tells us that the series converges.

Case 2: Behavior of the Function $n^2 e^{-n}$ as $n \to \infty$

If you are asking whether the function $n^2 e^{-n}$ converges to a limit as $n \to \infty$, we can analyze the limit:

$\lim_{n \to \infty} n^2 e^{-n}$

Since $e^{-n}$ decays exponentially as $n$ increases, it will dominate the polynomial growth of $n^2$. Thus, we expect:

$\lim_{n \to \infty} n^2 e^{-n} = 0$

This suggests that the function approaches zero as $n$ becomes large.

Conclusion

1. If it's a series: The series $\sum_{n=1}^{\infty} n^2 e^{-n}$ converges (based on the ratio test).
2. If it's the function: The function $n^2 e^{-n}$ tends to zero as $n \to \infty$.

Would you like further clarification or any additional details?

5 Related Questions:

1. What is the difference between convergence and divergence in a series?
2. How does the ratio test work in determining convergence?
3. What are other tests used to determine the convergence of series?
4. How do exponential and polynomial functions behave as $n \to \infty$?
5. What happens when you apply the limit comparison test to series?

Tip: When dealing with exponential decay like $e^{-n}$, it often dominates polynomial growth, which is why functions like $n^k e^{-n}$ tend to zero as $n \to \infty$.