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A power series is a series of the form \[ \sum_{n=0}^{\infty} c_n (x - a)^n \] where \( c_n \) are coefficients, \( x \) is a variable, and \( a \) is the center of the series. In the given power series \[ \sum_{n=1}^{\infty} \frac{1}{4^{n} \sqrt{n}} x^{n}, \] the coefficients are \( \frac{1}{4^n \sqrt{n}} \), and the series is centered at zero, making it a special case of a power series called a **power series centered at zero** or a **Maclaurin series**.Power series are crucial because they help represent functions as infinite polynomials, making them easier to work with under certain conditions. The convergence of a power series depends on the value of \( x \); specifically, on how far \( x \) is from the center \( a \).The concept of convergence leads us to explore tests like the **Ratio Test**, **p-Series Test**, and **Alternating Series Test** to determine where a power series converges. This is known as finding the **interval of convergence**.

The Ratio Test is a powerful tool for determining the convergence of a series, particularly handy for power series. To use the Ratio Test, we examine the ratio of the absolute values of consecutive terms in the series. Specifically, for a series \(\sum a_n\), the Ratio Test involves computing \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]Here's how it works:

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

In our original exercise, after applying the Ratio Test, we find \( L = \left| \frac{x}{4} \right|. \)Thus, the series converges when \( \left| \frac{x}{4} \right| < 1 \), which simplifies to the interval \(-4 < x < 4\). This interval is essential for identifying where the series behaves nicely, meaning it sums to a finite value.

The p-series test is a method used to determine the convergence of a series of the form \[ \sum \frac{1}{n^p} \]where \( p \) is a positive constant. For p-series:

• The series converges if \( p > 1 \).
• The series diverges if \( p \leq 1 \).

In the exercise step involving the endpoint \( x = 4 \), the series becomes \[ \sum_{n=1}^{\infty} \frac{1}{4^n \sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}. \]Here, \( p = 3/2 \), which is greater than 1, thus signaling convergence. Recognizing a p-series allows us to quickly conclude about the series' behavior, providing a straightforward solution for specific scenarios.

The Alternating Series Test is used to determine the convergence of series whose terms alternate in sign, often written as \[ \sum (-1)^n a_n. \]For convergence using this test, the series must satisfy:

• The absolute value of the terms \( a_n \) must decrease monotonically, that is, \( a_{n+1} \leq a_n \).
• The limit of the terms as \( n \to \infty \) must be zero, \( \lim_{n \to \infty} a_n = 0 \).

In our original exercise for \( x = -4 \), the series \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{4^n \sqrt{n}} \]is alternating. Verifying the conditions:- The terms decrease because \( \frac{1}{4^n \sqrt{n}} > \frac{1}{4^{n+1} \sqrt{n+1}} \).- The limit of each term is zero as \( n \to \infty \).Thus, the Alternating Series Test confirms convergence, solidifying our understanding that the series converges at this endpoint as well.