For this series, find the radius of convergence and write it as a geometric series and give a formula if $x>3$
$$\sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n$$
Now finding the radius of convergence wasn't too difficult and I'll save you guys the trouble of doing it because the interval of convergence is $ -1 = 3 – 4< x < 4 + 3 = 7$ and so the radius of convergence is 2
The second part confuses me because I don't understand (if it is even possible) to convert a power series to a geometric series.
Best Answer
For this particular function, shouldn't one note that $$ \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n = \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n}}(x-3)^n = \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x-3}{2}\right)^{n}$$ which is a geometric series!