Question

How do you find the interval of convergence for a power series?

Answer 1

In the above question you were asked to find the interval of convergence for a power series. For solving this problem you will need the ratio test. The ratio test states that if L < 1 then the series is converging and if L > 1 then it is divergent. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given question we have to find the interval of convergence for a power series. We will solve this problem using the ratio test
By Ratio Test,
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } |\dfrac{{{a_n} + 1}}{{{a_n}}}| = \mathop {\lim }\limits_{n \to \infty } |\dfrac{{{x^{n + 1}}}}{{n + 1}}.\dfrac{n}{{{x^n}}}| = |x|\mathop {\lim }\limits_{x \to \infty } \dfrac{n}{{n + 1}}$
$= |x|.1 = |x| < 1 \Rightarrow - 1 < x < 1$
It means that the power series converges on (-1, 1).
Now, we will check if the power series converges at x = -1 and x = 1.
The power series becomes alternating harmonic series when x = -1
$=\sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{n}}$
which is convergent. So, x = 1 should be considered.
The power series becomes the harmonic series when x = 1
$=\sum\limits_{n = 0}^\infty {\dfrac{1}{n}}$
which is divergent. So, x = 1 should not be considered.

Therefore, the interval of convergence is [-1, 1).

Note:
In the above solution we checked for two endpoints of x. For x = -1 the series was convergent and for x = 1 the series was divergent. Because of which we give the opening square “[” bracket for -1 and closed circular “)” bracket for 1. Also, as in integer -1 lies before 1 which is why we put -1 before 1 in the interval [-1, 1).