91Ó°ÊÓ

The radius of convergence is a key concept when dealing with power series like the one in our exercise, \(\sum C_n x^n\). Understanding this concept helps us determine over what interval the series will actually converge. The radius of convergence, denoted as \(R\), dictates that the series will converge for \(|x| < R\). To find \(R\), you can use the formula:\[\frac{1}{R} = \limsup_{n \to \infty} |C_n|^{1/n}\]In the example provided, each term \(C_n = \frac{(-1)^n}{n^2}\) leads us to calculate that:

• \(|C_n| = \frac{1}{n^2}\), giving us a simplicity as the terms become very small as \(n\) increases.
• \(\lim_{n \to \infty} n^{2/n} \to 1\), simplifying further calculations.

Thus, by applying the formula, we see that the radius of convergence is 1, meaning our series converges for values of \(x\) within \(-1 < x < 1\). However, our example also converges at two boundary points, \(x = 1\) and \(x = -1\), making it particularly interesting.

The Alternating Series Test is a method used to determine convergence for series that alternate in sign. It's particularly handy for the series we have when evaluated at \(x = 1\): \(\sum \frac{(-1)^n}{n^2}\).For a series \(\sum (-1)^n a_n\) to converge using this test, two conditions must be met:

• The absolute values of the terms \(a_n\) must be decreasing, meaning each term is smaller than the previous one.
• The terms must tend to zero as \(n\) approaches infinity, \(a_n \to 0\).

In our example, both these conditions are satisfied:

• The terms \(\frac{1}{n^2}\) get smaller as \(n\) increases, fulfilling the decreasing condition.
• Additionally, \(\lim_{n \to \infty} \frac{1}{n^2} = 0\), meeting the second criterion of the test.

Thus, by applying the Alternating Series Test, we can confidently say the series converges at \(x = 1\). This test makes analyzing series with alternating terms very manageable.

A p-series is often used when examining convergence in series like \(\sum\frac{1}{n^p}\). This type of series behaves differently depending on the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

In our exercise, when looking at \(x = -1\), the series simplifies to \(\sum \frac{1}{n^2}\), a classic p-series with \(p = 2\). Because \(2 > 1\), it definitely converges.To see why, consider the terms \(\frac{1}{n^2}\). As \(n\) increases, these terms rapidly decrease, closing in on zero. This characteristic trait is why p-series are so helpful in determining convergence: they provide a straightforward method based on \(p\).Understanding p-series serves as a quick way to determine whether the series will sum to a finite number or continue indefinitely. This power makes it a valuable tool when handling series with exponential-like behavior.