Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{\sqrt{n}}$$

Short answer

The interval of convergence is \((-1, 1]\).

Step-by-step solution

Identify the power series

Given the power series is \[\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{\sqrt{n}}\].

Use the Ratio Test

Apply the Ratio Test for absolute convergence. Consider \[a_{n} = \frac{(-1)^{n} x^{n}}{\sqrt{n}}\]and find\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right|.\]

Simplify the Ratio

Calculate\[\left| \frac{a_{n+1}}{a_{n}} \right| = \left| \frac{(-1)^{n+1} x^{n+1}}{\sqrt{n+1}} \cdot \frac{\sqrt{n}}{(-1)^{n} x^{n}} \right| = \left| x \right| \cdot \left| \frac{\sqrt{n}}{\sqrt{n+1}} \right|.\]

Take the Limit

Evaluate\[\lim_{n \to \infty} \left| x \right| \cdot \left| \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \left| x \right| \cdot \lim_{n \to \infty} \sqrt{\frac{n}{n+1}}.\]

Solve the Limit

As \( n \to \infty \), we know\[\sqrt{\frac{n}{n+1}} \to 1.\]Thus, the limit becomes\[\left| x \right|.\]

Apply the Ratio Test

For absolute convergence, the Ratio Test requires\[\left| x \right| < 1.\]So, the radius of convergence is 1.

Check the Endpoint \( x = 1 \)

Substitute \( x = 1 \): \[\sum_{n=1}^{\infty} \frac{(-1)^{n} (1)^{n}}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}.\] This is an alternating series with terms \(\frac{1}{\sqrt{n}}\) which decrease and tend to 0. By the Alternating Series Test, the series converges at \( x = 1 \).

Check the Endpoint \( x = -1 \)

Substitute \( x = -1 \): \[ \sum_{n=1}^{\infty} \frac{(-1)^{n} (-1)^{n}}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.\] This is the p-series with \( p = \frac{1}{2} \), which diverges (since \( p \le 1 \)). Therefore, the series diverges at \( x = -1 \).

Key concepts

Power Series

A power series is an infinite series of the form \(\sum_{n=0}^{\infty} a_n (x-c)^n\). Here, \(a_n\) is the coefficient of the \(n\)-th term and \(c\) is the center of the series. This form allows us to represent functions as sums of infinitely many polynomial terms.
A key feature of power series is their interval of convergence, within which the series converges to a function. Convergence means the series approaches a specific value as more terms are added. Outside this interval, the series may diverge, meaning it does not settle to any particular value.

Ratio Test

The Ratio Test helps determine whether an infinite series converges. For a series \(\sum_{n=0}^{\infty}a_n\), if the limit
\(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\) exists, we have:

• If \(L < 1\), the series converges absolutely.


• If \(L > 1\), the series diverges.


• If \(L = 1\), the test is inconclusive.

In the given exercise, applying the ratio test:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| x \right| \cdot \lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = \left| x \right|\]
This simplifies to \(|x|\), showing absolute convergence for \(\left| x \right| < 1\) with radius 1.

Alternating Series Test

The Alternating Series Test determines the convergence of series with terms that alternate in sign, such as \(\sum_{n=1}^{\infty}(-1)^n a_n\). For the series to converge, two conditions must be met:

• The absolute value of the terms, \(a_n\), must decrease monotonically, that is, \(|a_{n+1}| \leq |a_n|\).


• The limit of the terms must be zero as \(n\) approaches infinity, \(\lim_{n \to \infty} a_n = 0\).

For \(x = 1\) in our exercise, the series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\). This series meets both conditions:
\(|a_n| = \frac{1}{\sqrt{n}}\) decreases, and
\(\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\). Hence, the series converges at \(x = 1\).

P-Series

A p-series is of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a positive constant. The convergence of a p-series depends entirely on the value of \(p\):

• If \(p > 1\), the series converges.


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

When testing convergence at \(x = -1\) in the given problem, the series transforms into \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\), a p-series with \(p = \frac{1}{2}\). Since \(p = \frac{1}{2} \le 1\), the series diverges at \(-1\).