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

Answer 1

(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