Question

Consider the series [infinity] (-2)ⁿxⁿ ∑ ------------------ n=0 (n + 1)(n + 2) Find the interval of convergence of this series. FORMATTING: Your interval must be of one of the forms [a,b],(a,b],[a,b) or (a,b). Use appropriate interval notation, including the round or square parentheses to indicate whether the corresponding endpoint, a or b, is excluded or included in the interval. Interval:

Answer 1

To solve this problem, we will apply the ratio test to find the radius of convergence. We'll then check the endpoints separately.

The ratio test uses the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term of the series. If this limit (let's call it L) exists, the interval of convergence of the series will be |x-a| < 1/L. Here, a is 0 because our series is centered at 0 (it's a power series in xⁿ).

First, let's express the nth and (n+1)th terms. The nth term, cn, is (-2)ⁿ / (n + 1)(n + 2), and the (n+1)th term, cn+1, is (-2)ⁿ⁺¹ / (n + 2)(n + 3).

Taking the absolute value of the ratio cn+1/cn gives us |-2(n + 1)/(n + 3)|. As n goes to infinity, this limit simplifies to |-2|.

Our radius of convergence (R) then becomes 1/|-2|, which simplifies to 1/2. The series will converge absolutely for |x-0| < 1/2, which is the same as -1/2 < x < 1/2.

Now, we need to check convergence at the endpoints, x = 0±R = 0±1/2 = ±1/2. When we substitute these values into our original series, it simplifies to ±1 / (n + 1)(n + 2). The series 1 / n(n + 1) is a p-series with p = 2. Since p > 1, this series converges by the p-series test.

Including both endpoints, the interval of convergence of this series is therefore [-1/2, 1/2].