Question

In Exercises \(3-6,\) find the radius of convergence of the power series. $$ \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n}}{n+1} $$

Short answer

The radius of convergence of the power series \( \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n}}{n+1} \) is \( \infty \).

Step-by-step solution

Formulate the ratio of the (n+1)th term to the nth term

We have been given the power series \( \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n}}{n+1} \). Here, the nth term is \( (-1)^{n} \frac{x^{n}}{n+1} \). Therefore, the (n+1)th term is \( (-1)^{n+1} \frac{x^{n+1}}{(n+2)} \). We then form the ratio of the (n+1)th term to the nth term, which is \( \frac{(-1)^{n+1} \frac{x^{n+1}}{(n+2)}}{(-1)^{n} \frac{x^{n}}{n+1}} \). This simplifies to \( \frac{x(-1)}{(n+2)} \).

Take the absolute value and limit as n approaches infinity

Now we take the absolute value of the ratio from Step 1: \( \left|\frac{x(-1)}{(n+2)}\right| = |x| \frac{1}{(n+2)} \). Then, we take the limit of this absolute value as n approaches infinity to determine its convergence. This limit is \(\lim_{{n \to \infty}} \frac{|x|}{(n+2)} = 0 \). Thus the series converges for all x, and the radius of convergence is therefore infinite.

Conclude the radius of convergence

Because the limit is equal to 0 (less than 1), the series converges for all x according to the Ratio Test. This indicates that the radius of convergence, the distance from the center to the boundary of the interval of convergence, is infinity. The radius of convergence of the series is therefore \( \infty \).