I've been stuck on the following question for a while now, where i have to use the geometric series to expand the following function into a power series at the given centre, and find the radius of convergence, which is;
$$\frac1{1+x}\text{ about }x_0=2$$
I have managed to take 1/3 out and then have attempted to use the ratio test, which i think is the right approach but i can't seem to get a valid answer.
$$\begin{array}{rcl} \displaystyle \frac1{1+x} &=& \displaystyle \frac1{3+(x-2)} \\ &=& \displaystyle \frac13 \dfrac1{1+\frac{x-2}3} \\ &=& \displaystyle \frac13 \sum_{n=0}^\infty \left(-\frac{x-2}3\right)^n \\ &=& \displaystyle \frac13 \sum_{n=0}^\infty \left(-\frac13\right)^n (x-2)^n \\ \end{array}$$
We need to rewrite that fraction to look like the sum of a geometric series, i.e., $\frac{1}{1-r}$, and we need for $r$ to somehow be in terms of $(x-2)$. Thus:
$$\begin{align} \frac{1}{1+x} &= \frac{1}{3+x-2}\\ &= \frac{1}{3--(x-2)}\\ &=\frac33\cdot\frac{\frac13}{1-\left(-\frac{x-2}{3}\right)}\\ &=\frac13\cdot\frac{1}{1-\left(-\frac{x-2}{3}\right)} \end{align}$$
Now it's in the right form, so we can write it as a geometric series:
$$\frac1{1+x}=\frac13\sum_{n=0}^\infty \left(-\frac{x-2}{3}\right)^n=\frac13\sum_{n=0}^\infty \left(\frac{-1}{3}\right)^n(x-2)^n$$
Does that make sense?
