Find the power series representation and interval of convergence for the function $ \ f(x)=\frac{3}{2+x} \ $
Answer:
$f(x)=\frac{3}{x+2}=\frac{3}{2} (1+\frac{x}{2})^{-1}=\frac{3}{2} (1-\frac{x}{2}+(\frac{x}{2})^2-(\frac{x}{2})^3+..........) \ =\sum_{n=0}^{\infty} \frac{3}{2} (-1)^n (\frac{x}{2})^n $
The series converges if $ \ |\frac{x}{2}|<1 \ \Rightarrow |x|<2 \ $
Thus the interval of convergence is $ -2<x<2 \ $
Am I right so far?
Rewrite the series as $\; \sum_{n\ge 0}(-1)^n 16^n x^{4n+5}= x^5\sum_{n\ge 0}(-1)^n 16^n(x^4)^n$.
The radius of convergence of the power series (setting $x^4=u$) is given by $$\frac1R=\limsup_n\,\bigl(16^n\bigr)^{\tfrac1n}=16,$$ so the given power series converges if
$$ x^4<\frac1{16}\iff |x|<\frac12.$$