Find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x. a) Find the values of x for which the given geometric series converges. b) Find the sum of the series sum_{n=0}^infty(-frac12)^n(x-5)^n

(a) Consider the given geometric series:
$\sum _{n=0}^{\mathrm{\infty }}(-\frac{1}{2}{)}^n(x-5{)}^n$
Now, to find the value of x for which the series converges, apply ratio test by finding the following limit:
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$ where $a_n=(-\frac{1}{2}{)}^n(x-5{)}^n$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{(-\frac{1}{2}{)}^{n+1}(x-5{)}^{n+1}}{(-\frac{1}{2}{)}^n(x-5{)}^n}|$
$=\underset{n\to \mathrm{\infty }}{lim}|(-\frac{1}{2})(x-5)|$
$=|\frac{5-x}{2}|$
Now, for the series to be convergent, the limit should be less than 1, that is:
$L<1$
$|\frac{5-x}{2}|<1$
$-1<\frac{5-x}{2}<1$
$-2<5-x<2$
$-7<-x<-3$
Now, check the convergence at 7 and 3:
At x=7
$\sum _{n=0}^{\mathrm{\infty }}(-\frac{1}{2}{)}^n(7-5{)}^n=\sum _{n=0}^{\mathrm{\infty }}(-\frac{1}{2}{)}^n{2}^n=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\to \mathrm{\infty }$
Which is divergent
At x=3
$\sum _{n=0}^{\mathrm{\infty }}(-\frac{1}{2}{)}^n(3-5{)}^n=\sum _{n=0}^{\mathrm{\infty }}(-\frac{1}{2}{)}^n(-2{)}^n=\sum _{n=0}^{\mathrm{\infty }}(1{)}^n\to \mathrm{\infty }$
Which is divergent
Therefore, the required values of x for which the series converges is:
(b) Now, since, given series is a geometric series with:
$a=1,r=(-\frac{1}{2})(x-5)=(\frac{5-x}{2})$
Therefore, the sum of this infinite series will be given by:
$S=\frac{a}{1-r}$
$=\frac{1}{(\frac{5-x}{2})}$
$=\frac{2}{5-x}$
Thus, required sum of series is:
$S=\frac{2}{5-x}$ Where 3