Find a power series for function

Question

I'm having some difficulty with this problem even while noting the hint.

I expressed the function as $(1/2)\frac{1}{1-(-3x/2)}$ and then thought I would work with $1/2$ of the infinite sum of $(-3/2)^n x^n$ but I'm not sure if I'm heading in the right direction. Does that sound right?

Find a power series for $f(x)=\dfrac1{2+3x}$. Hint: Use $\dfrac1{1-x}=\displaystyle\sum_{n_0}^\infty x^n$.

Answer 1

You essentially solved it; just make the substitution $$ f(x) = \frac{1}{2+3x} = \frac{1}{2}\left(\frac{1}{1-\left(\frac{-3x}{2}\right)} \right) = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{-3x}{2}\right)^n. $$ Note that $1/(1-x)$ converges when $|x| < 1$, though, so in this problem the radius of convergence is $$ \left|\frac{-3x}{2}\right| < 1 \iff \left|x\right| < \frac{2}{3}. $$

Answer 2

$$\dfrac{1}{2+3x}=\dfrac{1}{2}\cdot\dfrac{1}{1+\frac{3}{2}x}=\dfrac{1}{2}\cdot\sum_{n=0}^{\infty}(-\dfrac{3x}{2})^n$$

Because you see that $$\dfrac{1}{1+\frac{3}{2}x}=\dfrac{1}{1-t}=\sum_{n=0}^{\infty}t^n,$$ where $t=-\dfrac{3x}{2}$.