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## Homework Statement

[tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(x-1)^{2n+1}}{2n+1}[/tex]

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

[tex]\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}[/tex]

[tex](x-1)^{2}\lim_{n\rightarrow\infty}\frac{2n+1}{2n+3}=(x-1)^{2}[/tex]

[tex](x-1)^{2}<1[/tex]

[tex]\sqrt{(x-1)^{2}}<1[/tex]

[tex]\vert x-1\vert<1[/tex]

[tex]0<x<2[/tex]

so my radius of convergence is 1

then plugging in x=0 and x=2

i get [tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(-1)^{2n+1}}{2n+1}[/tex] and

[tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}(1)^{2n+1}}{2n+1}[/tex]

i found that at both x=0 and x=2 the series converges conditionally

so then my interval of convergence is [tex]0\leq x \leq 2[/tex] 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?