(a) Consider \(\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}\).

Using the method of differences, the terms cancel out in a pattern, leading to the final result.

(b) For convergence, \(e^{nx} \to 0\) as \(n \to \infty\), which occurs when \(x < 0\). The sum to infinity is \(e^x - e^{2x}\).

(c) Using the laws of logarithms, \(\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n} (rx + \ln(e^x - 1)^2)\).

Applying the formula \(\sum_{r=1}^{n} r = \frac{1}{2} n(n+1)\), we get \(\frac{1}{2} xn(n+1) + n \ln(e^x - 1)^2\).