Question

Let $S_{k}, k =1,2, \ldots, 100$, denote the sum of the infinite geometric series whose first term is $\frac{ k -1}{ k !}$ and the common ratio is $\frac{1}{ k } .$ Then the value of $\frac{100^{2}}{100 !}+\displaystyle\sum_{ k =1}^{100}\left|\left( k ^{2}-3 k +1\right) S _{ k }\right|$ is _____

Answer 1

$S_{k}=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}$
$=\displaystyle\sum_{k=2}^{100}\left|\left(k^{2}-3 k+1\right) \frac{1}{(k-1) !}\right|$
$=\displaystyle\sum_{k=2}^{100}\left|\frac{(k-1)^{2}-k}{(k-1) !}\right|$
$=\sum\left|\frac{k-1}{(k-2) !}-\frac{k}{(k-1) !}\right|$
$=\left|\frac{2}{1 !}-\frac{3}{2 !}+\right| \frac{3}{2 !}-\frac{4}{3 !} \mid+\cdots$
$=\frac{2}{1 !}-\frac{1}{0 !}+\frac{2}{1 !}-\frac{3}{2 !}+\frac{3}{2 !}-\frac{4}{3 !}+\cdots+\frac{99}{98 !}-\frac{100}{99 !}$
$=3-\frac{100}{99 !}$