Find the interval of convergence of the power series.(Be sure to include a check for convergence at the endpoints of the interval. )

Question

Find the interval of convergence of the power series.(Be sure to include a check for convergence at the endpoints of the interval. ) ​  ​

Quizplus · Accepted Answer

The interval of convergence for a power series $\sum_{n=0}^{\infty} c_n(x-a)^n$ can be found by applying the ratio test: 
$$\lim_{n 	o \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}ight| = \lim_{n 	o \infty} |c_{n+1}| |x-a|$$
This limit exists if it is finite, so we want to find values of $x$ such that the limit is less than 1.
For our power series $\sum_{n=1}^{\infty} \frac{(-x)^n}{n}$, we have $c_n = \frac{(-1)^n}{n}$. Applying the ratio test gives:
$$\lim_{n 	o \infty} \left|\frac{c_{n+1}x^{n+1}}{c_n x^n}ight| = \lim_{n 	o \infty} \frac{n}{n+1} |x| = |x|$$
This limit is less than 1 if $|x| < 1$, so the interval of convergence is $-1 < x < 1$.
To check the endpoints, we can test whether the series converges or diverges when $x = -1$ and $x = 1$.
When $x = -1$, we have:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n} = - \ln 2$$
This is a convergent alternating series, so the power series converges at $x = -1$.
When $x = 1$, we have:
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
This is the harmonic series, which is known to diverge. Therefore, the power series diverges at $x = 1$.
Therefore, the interval of convergence is $-1 \leq x < 1$, and the best choice is B.

Find the Interval of Convergence of the Power Series