Question

Find the power series representation for the function and determine the interval of convergence

f(x)=2/(3-x)

Answer 1

Find the nth derivative:
\[D \left( \frac{2}{3-x}\right)=\frac{2}{(3-x)^2}; D^2 \left( \frac{2}{(3-x)}\right)=\frac{2(2)}{(3-x)^3} \]
\[\implies D^n \left( \frac{2}{3-x}\right) \left( \frac{2n!}{(3-x)^{n+1}}\right)\]
Evaluating this at x=0 gives:
\[\frac{2n!}{3^{n+1}}\]
So:
\[f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x)(x-a)^n}{n!}\]
I'm assuming you want the McLaurin series where a=0 so:
\[f(x)=\sum_{n=0}^{\infty} \frac{2x^n}{3^{n+1}}\]
Then apply the ratio test:
\[\lim_{k \rightarrow \infty}\left| \frac{a_{k+1}}{a_k} \right|<|1|\]
This gives you the interval. Then plug in the end points for x to find if its less than or less than or equal to.