The given series is:

$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^2 \cdot 2^n}{n!}$

To determine if this series is absolutely convergent, conditionally convergent, or divergent, let's proceed by analyzing it step-by-step.

Step 1: Check for Absolute Convergence

To check for absolute convergence, we examine the series with the absolute value of each term:

$\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n^2 \cdot 2^n}{n!} \right| = \sum_{n=1}^{\infty} \frac{n^2 \cdot 2^n}{n!}$

Step 2: Use the Ratio Test

The ratio test is often useful for series involving factorials. Let's apply it to the absolute series $\sum_{n=1}^{\infty} \frac{n^2 \cdot 2^n}{n!}$.

The ratio test states that for a series $\sum a_n$, we examine:

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

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

Let's calculate $a_n = \frac{n^2 \cdot 2^n}{n!}$ and find $a_{n+1}$:

$a_{n+1} = \frac{(n+1)^2 \cdot 2^{n+1}}{(n+1)!}$

Now, compute $\frac{a_{n+1}}{a_n}$:

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

As $n \to \infty$, $\frac{2(n+1)}{n^2} \to 0$, so $L = 0$.

Since $L < 1$, the series $\sum_{n=1}^{\infty} \frac{n^2 \cdot 2^n}{n!}$ converges absolutely.

Conclusion

Because the series converges absolutely, it also converges.

Answer

The series is absolutely convergent.

Would you like more details or have questions about any step?

Here are 5 related questions to further explore the topic:

1. How does conditional convergence differ from absolute convergence?
2. What other tests can be used for determining convergence of series involving factorials?
3. Can the alternating series test be applied to this series directly?
4. How does the ratio test work when applied to power series?
5. What are the implications of a series being absolutely convergent in real analysis?

Tip: When a series includes factorial terms, the ratio test is often a quick and effective method to test for convergence.