009C Sample Final 2, Problem 7

(a) Consider the function  $f(x)={\bigg (}1-{\frac {1}{2}}x{\bigg )}^{-2}.$  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

Foundations:

Solution:

(a)

Step 1:

Step 2:

(b)

Step 1:
The Maclaurin series of  ${\frac {1}{(1-x)^{2}}}$  is
$\sum _{n=0}^{\infty }(n+1)x^{n}.$
So, the Maclaurin series of  ${\frac {1}{(1-{\frac {1}{2}}x)^{2}}}$  is
$\sum _{n=0}^{\infty }(n+1){\bigg (}{\frac {1}{2}}x{\bigg )}^{n}=\sum _{n=0}^{\infty }{\frac {(n+1)x^{n}}{2^{n}}}.$

Step 2:
Now, we use the Ratio Test to determine the radius of convergence of this power series.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+2)x^{n+1}}{2^{n+1}}}{\frac {2^{n}}{(n+1)x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {|x|}{2}}{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}\lim _{n\rightarrow \infty }{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}.}\end{array}}$
Now, the Ratio Test says this series converges if  ${\frac {|x|}{2}}<1.$  So,  $|x|<2.$
Hence, the radius of convergence is  $R=2.$

Final Answer:
(a)
(b)    The radius of convergence is  $R=2.$

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