7. (a) Find the interval of convergence for $f(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n}(x-1)^{n}}{5^{n}}$.
(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 \(\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L\), then the series \(\sum_{n=0}^{\infty} a_n\) 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 \(\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1}(x-1)^{n+1}}{5^{n+1}} \div \frac{(-1)^{n}(x-1)^{n}}{5^{n}} = -\frac{(x-1)}{5}\).

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

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 \(f(4) = \sum_{n=0}^{\infty} \frac{(-1)^{n}(4-1)^{n}}{5^{n}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}3^{n}}{5^{n}}\).

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 \(f(-2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}(-2-1)^{n}}{5^{n}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}(-3)^{n}}{5^{n}}\).

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 \(f(-5) = \sum_{n=0}^{\infty} \frac{(-1)^{n}(-5-1)^{n}}{5^{n}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}(-6)^{n}}{5^{n}}\).

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.