7. (a) Find the interval of convergence for $f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(x-1{)}^n}{5^n}$.
(b) Find the exact value of $f(4)$, if possible.
(c) Find the exact value of $f(-2)$, if possible.
(d) Find the exact value of $f(-5)$, if possible.

Step 1 ：(a) We use the Ratio Test to find the interval of convergence. The Ratio Test states that if $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=L$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges if $L<1$ and diverges if $L>1$. If $L=1$, the test is inconclusive. Here, $a_n=\frac{(-1{)}^n(x-1{)}^n}{5^n}$.

Step 2 ：We find $\frac{a_{n+1}}{a_n}=\frac{(-1{)}^{n+1}(x-1{)}^{n+1}}{5^{n+1}}÷\frac{(-1{)}^n(x-1{)}^n}{5^n}=-\frac{(x-1)}{5}$.

Step 3 ：Taking the absolute value, we get $|-\frac{(x-1)}{5}|=\frac{|x-1|}{5}$.

Step 4 ：We set $\frac{|x-1|}{5}<1$ and solve for $x$ to get the interval of convergence. This gives us $-4<x<6$.

Step 5 ：(b) To find the exact value of $f(4)$, we substitute $x=4$ into the series. This gives us $f(4)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(4-1{)}^n}{5^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{3}^n}{5^n}$.

Step 6 ：This is a geometric series with first term $a=1$ and common ratio $r=-\frac{3}{5}$. The sum of a geometric series is $\frac{a}{1-r}$, so $f(4)=\frac{1}{1-(-\frac{3}{5})}=\frac{5}{8}$.

Step 7 ：(c) To find the exact value of $f(-2)$, we substitute $x=-2$ into the series. This gives us $f(-2)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(-2-1{)}^n}{5^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(-3{)}^n}{5^n}$.

Step 8 ：This is a geometric series with first term $a=1$ and common ratio $r=\frac{3}{5}$. The sum of a geometric series is $\frac{a}{1-r}$, so $f(-2)=\frac{1}{1-\frac{3}{5}}=\frac{5}{2}$.

Step 9 ：(d) To find the exact value of $f(-5)$, we substitute $x=-5$ into the series. This gives us $f(-5)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(-5-1{)}^n}{5^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(-6{)}^n}{5^n}$.

Step 10 ：However, $-5$ is not in the interval of convergence $-4<x<6$, so the series does not converge and $f(-5)$ is undefined.