Question

# 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

Series
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
$$\sum_{n=0}^\infty(-\frac12)^n(x-5)^n$$

2020-11-09

(a) Consider the given geometric series:
$$\sum_{n=0}^\infty(-\frac12)^n(x-5)^n$$
Now, to find the value of x for which the series converges, apply ratio test by finding the following limit:
$$L=\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|$$ where $$a_n=(-\frac12)^n(x-5)^n$$
$$=\lim_{n\rightarrow\infty}|\frac{(-\frac12)^{n+1}(x-5)^{n+1}}{(-\frac12)^n(x-5)^n}|$$
$$=\lim_{n\rightarrow\infty}|(-\frac12)(x-5)|$$
$$=|\frac{5-x}{2}|$$
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
$$\sum_{n=0}^\infty(-\frac12)^n(7-5)^n=\sum_{n=0}^\infty(-\frac12)^n2^n=\sum_{n=0}^\infty(-1)^n\rightarrow\infty$$
Which is divergent
At x=3
$$\sum_{n=0}^\infty(-\frac12)^n(3-5)^n=\sum_{n=0}^\infty(-\frac12)^n(-2)^n=\sum_{n=0}^\infty(1)^n\rightarrow\infty$$
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:
$$a=1,r=(-\frac12)(x-5)=(\frac{5-x}{2})$$
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