Question

An alternating series \(\sum_{n=i}^{\infty} a_{n}\) is given. (a) Determine if the series converges or diverges. (b) Determine if \(\sum_{n=0}^{\infty}\left|a_{n}\right|\) converges or diverges. (c) If \(\sum_{n=0}^{\infty} a_{n}\) converges, determine if the convergence is conditional or absolute. $$\sum_{n=0}^{\infty} \frac{(-1)^{n} n^{2}}{n !}$$

Short answer

The series converges absolutely as both the series and its absolute series converge.

Step-by-step solution

Analyze the Alternating Series for Convergence

The given series is \(\sum_{n=0}^{\infty} \frac{(-1)^{n} n^{2}}{n!}\). According to the Alternating Series Test, a series \(\sum (-1)^n a_n\) converges if the absolute value of the terms \(a_n\) decreases monotonically (i.e., \(a_{n+1} \leq a_n\)) and the limit \(\lim_{n \to \infty} a_n = 0\). Here, \(a_n = \frac{n^2}{n!}\). As \(n\) increases, \(n!\) grows much faster than \(n^2\), so \(a_n\) decreases and tends to zero. Thus, the given series converges.

Determine Convergence of the Absolute Series

Consider the series \(\sum_{n=0}^{\infty} \left|a_n\right| = \sum_{n=0}^{\infty} \frac{n^2}{n!}\). Using the Ratio Test, we find the ratio \(r_n = \frac{(n+1)^2}{(n+1)!} \cdot \frac{n!}{n^2} = \frac{(n+1)^2}{(n+1)n^2} = \frac{n+1}{n^2}\). As \(n\to\infty\), \(r_n\to 0\), which is less than 1, indicating convergence of \(\sum_{n=0}^{\infty} \left|a_n\right|\). Therefore, the absolute series converges.

Classify the Convergence Type

Since the original series \(\sum_{n=0}^{\infty} \frac{(-1)^{n} n^{2}}{n!}\) converges and the absolute series \(\sum_{n=0}^{\infty} \left|a_n\right|\) also converges, the convergence is absolute. Absolute convergence implies that the series converges regardless of changes in the order of its terms.

Key concepts

Absolute Convergence

Absolute convergence is an important concept in the study of infinite series. It means that even when you take the absolute value of each term in the series and sum them up, the series still converges. This is a stronger condition than just plain convergence because it guarantees the series' stability even if we rearrange its terms.
For a series \( \sum a_n \), if \( \sum \left|a_n\right| \) converges, then \( \sum a_n \) is said to be absolutely convergent. Absolute convergence implies convergence, but not vice versa. In the given problem, both the series \( \sum \frac{(-1)^n n^2}{n!} \) and its absolute counterpart \( \sum \frac{n^2}{n!} \) converge, thus proving that our original series is absolutely convergent.
This concept is powerful because of the Absolute Convergence Theorem, which tells us that absolutely convergent series can be safely rearranged without affecting their sum. This is not always true for conditionally convergent series, where rearranging can lead to different sums.

Alternating Series Test

The Alternating Series Test is a nifty tool to determine whether an alternating series converges. An alternating series takes the form \( \sum (-1)^n a_n \), where the signs of terms change from positive to negative or vice versa.
To apply the Alternating Series Test, two conditions must be met:

• The absolute values of the terms \(a_n\) must decrease monotonically (i.e., each successive term is smaller or equal to the previous term).
• The sequence \( \lim_{n \to \infty} a_n = 0 \).

If both these conditions are satisfied, then the alternating series converges.
In our problem, the given series \( \sum_{n=0}^{\infty} \frac{(-1)^n n^2}{n!} \) does meet these criteria. The terms decrease because factorial \( n! \) grows much faster than the square \( n^2 \), and the limit condition is also satisfied. Thus, according to the Alternating Series Test, this series converges.

Ratio Test

The Ratio Test is a popular method for testing the convergence of series, particularly when terms involve factorials or exponential functions. To use the Ratio Test, we calculate the limit of the ratio of successive terms.
The test involves computing:
\[ r_n = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

• If \( 0 \leq r_n < 1 \), the series absolutely converges.
• If \( r_n > 1 \) or is infinite, the series diverges.
• If \( r_n = 1 \), the test is inconclusive.

In our given series \( \sum_{n=0}^{\infty} \frac{n^2}{n!} \), we applied the Ratio Test by comparing terms \( \frac{n^2}{n!} \) and found that the limit \( r_n \to 0 \) as \( n \to \infty \), implying convergence. This analysis showed that even the series' absolute value converges, confirming absolute convergence by the Ratio Test.