Science:Math Exam Resources/Courses/MATH101/April 2012/Question 05/Solution 1

We will use the ratio test:

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\frac {(x-1)^{n+1}}{2^{n+1}(n+3)}}{\frac {(x-1)^{n}}{2^{n}(n+2)}}}\right|\\&=\lim _{n\to \infty }{\frac {|x-1|(n+2)}{2\cdot (n+3)}}\\&=|{\frac {x-1}{2}}|\end{aligned}}$

We want this limit to be $<1$ so that the series converges absolutely. This means $|{\frac {x-1}{2}}|<1$ which means that the radius of convergence is $2$ and that $-1<x<3$. Now we have to test the endpoints.

At $x=3$ we get:

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

which converges by alternate series test, if we sum two consecutive terms we get:

${\frac {1}{n}}-{\frac {1}{n+1}}={\frac {1}{n(n+1)}}<{\frac {1}{n^{2}}}$

and the series $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ converges.

At $x=-1$ we get

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

which diverges because it is the tail of the harmonic series.

So, the interval of convergence is $(-1,3].$