(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