Originally Posted by

**sbhatnagar** $\displaystyle S_k$ is an infinite geometric series.

1.Sum of an infinite geometric series is given by $\displaystyle S=\frac{a}{1-r}$

therefore $\displaystyle S_k= \frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

$\displaystyle \implies S_k = \frac{1}{(k-1)!}$

2. Note That:

$\displaystyle \begin{align*} \sum_{k=1}^{100}\Big| \frac{k^2-3k+1}{(k-1)!}\Big|&= \sum_{k=1}^{100}\Big| \frac{(k-1)^2-k}{(k-1)!}\Big| \\ &=\sum_{k=1}^{100}\Big| \frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}\Big| \\ &= |-1|+|-1|+\Big| \frac{2}{1!}-\frac{3}{2!}\Big|+\Big| \frac{3}{2!}-\frac{4}{3!}\Big|+\cdots + \Big| \frac{99}{98!}-\frac{100}{99!}\Big| \\ &= 1+1+2-\frac{100}{99!} \\ &=4-\frac{100}{99!}\end{align*}$