# Check my solution to a power series/taylor series problem

## Homework Statement

$$\sum_{n=0}^{\infty}\frac{(-1)^{n}(x-1)^{2n+1}}{2n+1}$$

Find the series' radius and interval of convergence. Then identify the values of x for which the series converges absolutely and conditionally.

## Homework Equations

ratio test
absolute convergence test
alternating series test

## The Attempt at a Solution

$$\lim_{n\rightarrow\infty}\vert\frac{(x-1)^{2n+3}}{2n+3}\frac{2n+1}{(x-1)^{2n+1}}\vert=\lim_{n\rightarrow\infty}(x-1)^{2}\frac{2n+1}{2n+3}$$
$$(x-1)^{2}\lim_{n\rightarrow\infty}\frac{2n+1}{2n+3}=(x-1)^{2}$$
$$(x-1)^{2}<1$$
$$\sqrt{(x-1)^{2}}<1$$
$$\vert x-1\vert<1$$
$$0<x<2$$
so my radius of convergence is 1
then plugging in x=0 and x=2
i get $$\sum_{n=0}^{\infty}\frac{(-1)^{n}(-1)^{2n+1}}{2n+1}$$ and
$$\sum_{n=0}^{\infty}\frac{(-1)^{n}(1)^{2n+1}}{2n+1}$$
i found that at both x=0 and x=2 the series converges conditionally
so then my interval of convergence is $$0\leq x \leq 2$$ with absolute convergence on 0<x<2 and conditional convergence at x=0 and 2
i have a test tomorrow on sequences, infinite series, power series, and taylor series so im working on the problems in my book but this one is an even numbered problme so i cant check my solution in the back of the book
also i noticed that this series is the taylor series for arctanx centered at x=1, could i have used this to get the solution much quicker than what i did?

well at x=0 the sequence $$\frac{1}{2n+1}$$ is positive for all n, its decreasing because 2n+1 is an increasing sequence and the reciprocal of an increasing sequence is decreasing and the sequence goes to zero as n goes to infinity so by the alternating series test the series converges but it diverges when applying the absolute convergence test to the series at x=0 and limit comparison with 1/n