Answer
See explanation
Work Step by Step
a) $s_1 = \frac{1}{2}, s_2 = \frac{5}{6}, s_3 = \frac{23}{24}, s_4=\frac{119}{120}, s_n = \frac{(n+1)! - 1}{(n+1)!}$
b) This formula is true to n = 1, so suppose it's valid for n = k, $s_k = \frac{(k+1)! - 1}{(k+1)!}$, then taking n = k+1 we see that it's valid too, $s_{k+1} = s_k + \frac{k+1}{(k+2)!} = \frac{(k+1)! - 1}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{(k+2)! - k - 2 + k +1}{(k+2)!} = \frac{((k+1)+1)! - 1}{((k+1)+1)!}$
c) Using what we've found in b), $\lim_{k \to \infty} s_k =\lim_{k \to \infty} (1 - \frac{1}{(k+1)!}) = 1 $ because $\lim_{k \to \infty} (k+1)! = +\infty$
