Questions: Find R, the radius of convergence, and the open interval of convergence for: sum from n=1 to infinity of ((-1)^(n+1)(x+2)^n)/(n 2^n) R= Sigma Interval = What happens at the endpoints of the interval?

Solution Steps

To find the radius of convergence \( R \) for the given power series, we can use the ratio test. The ratio test involves taking the limit of the absolute value of the ratio of consecutive terms. For a series \(\sum a_n\), the radius of convergence \( R \) is given by \( R = \frac{1}{\limsup_{n \to \infty} |a_{n+1}/a_n|} \). Once \( R \) is found, the interval of convergence can be determined by checking the convergence at the endpoints separately.

The given power series is:

\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} \]

This is a power series of the form:

\[ \sum_{n=1}^{\infty} a_n (x - c)^n \]

where \( a_n = \frac{(-1)^{n+1}}{n 2^n} \) and \( c = -2 \).

To find the radius of convergence \( R \), we use the ratio test. The ratio test states that a series \(\sum a_n\) converges if:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]

For our series, we have:

\[ a_n = \frac{(-1)^{n+1}}{n 2^n} \]

\[ a_{n+1} = \frac{(-1)^{n+2}}{(n+1) 2^{n+1}} \]

Now, calculate the ratio:

\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+2}}{(n+1) 2^{n+1}}}{\frac{(-1)^{n+1}}{n 2^n}} \right| = \left| \frac{(-1)^{n+2} \cdot n \cdot 2^n}{(-1)^{n+1} \cdot (n+1) \cdot 2^{n+1}} \right| \]

Simplifying, we get:

\[ = \left| \frac{n}{(n+1) \cdot 2} \right| = \frac{n}{2(n+1)} \]

Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{n}{2(n+1)} = \lim_{n \to \infty} \frac{n}{2n + 2} = \lim_{n \to \infty} \frac{1}{2 + \frac{2}{n}} = \frac{1}{2} \]

The series converges when:

\[ \left| x + 2 \right| < R \]

From the ratio test, we have:

\[ \frac{1}{2} < 1 \]

Thus, the radius of convergence \( R \) is 2.

The interval of convergence is determined by:

\[ -2 - R < x < -2 + R \]

Substituting \( R = 2 \):

\[ -2 - 2 < x < -2 + 2 \]

\[ -4 < x < 0 \]

We need to check the convergence at the endpoints \( x = -4 \) and \( x = 0 \).

1. At \( x = -4 \):

   The series becomes:

   \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-2)^n}{n 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \]

   This is the harmonic series, which diverges.

2. At \( x = 0 \):

   The series becomes:

   \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2)^n}{n 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \]

   This is the alternating harmonic series, which converges.

Final Answer