Check my solution to a power series/taylor series problem

[miglo]

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 I am working on the problems in my book but this one is an even numbered problme so i can't 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?

 

[Dick]

That looks fine to me if you are clear on why x=0 and x=2 cases converge by the alternating series test. Knowing it's the series for arctan(x) doesn't help unless you know something about complex analysis (maybe). I think you are doing it the simplest way possible.

 

[miglo]

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
at x=2 the sequence is the same as the above so it converges by alternating series
but it fails the absolute convergence test in the same way the first one failed so at both points they only converge conditionally
ohh i see, i was hoping that the taylor series would make things quicker