# 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

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
$\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}\left(x-5{\right)}^{n}$
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Nathanael Webber

(a) Consider the given geometric series:
$\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}\left(x-5{\right)}^{n}$
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 ${a}_{n}=\left(-\frac{1}{2}{\right)}^{n}\left(x-5{\right)}^{n}$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{\left(-\frac{1}{2}{\right)}^{n+1}\left(x-5{\right)}^{n+1}}{\left(-\frac{1}{2}{\right)}^{n}\left(x-5{\right)}^{n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\left(-\frac{1}{2}\right)\left(x-5\right)|$
$=|\frac{5-x}{2}|$
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
$\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}\left(7-5{\right)}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}{2}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\to \mathrm{\infty }$
Which is divergent
At x=3
$\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}\left(3-5{\right)}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{1}{2}{\right)}^{n}\left(-2{\right)}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(1{\right)}^{n}\to \mathrm{\infty }$
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:
$a=1,r=\left(-\frac{1}{2}\right)\left(x-5\right)=\left(\frac{5-x}{2}\right)$
Therefore, the sum of this infinite series will be given by:
$S=\frac{a}{1-r}$
$=\frac{1}{\left(\frac{5-x}{2}\right)}$
$=\frac{2}{5-x}$
Thus, required sum of series is:
$S=\frac{2}{5-x}$ Where 3

Jeffrey Jordon