the value of $\frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_{k}|$ is

Question

Let $S_{k},$ where $k=1,2,3,\cdots \cdots ,100$ denote the sum of the infinite geometric series whose first term is $\displaystyle \frac{k-1}{k!}$ and the common ratio is $\displaystyle \frac{1}{k},$ then the value of $\displaystyle \frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_{k}|$ is

$\displaystyle S_{k} = \frac{a}{1-r} = \frac{(k-1)\cdot k}{k!\cdot (k-1)} = \frac{1}{(k-1)!}$

So $\displaystyle \sum^{100}_{k=1}\bigg|(k^2-3k+1)S_{k}\bigg| = \sum^{100}_{k=1}\bigg|\frac{(k-1)^2-k}{(k-1)!}\bigg| = \sum^{100}_{k=1}\bigg|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}\bigg| $

wan,t be able to go further, could some help me with this, thanks

Answer 1

Hint thats a telescpoic series thus only first and last terms survive so summation is $|1-\frac {100}{99!}| $ also starting series with $k=2$ makes sense as i dont know whether value of $(-1)!$ is defined

Answer 2

$$\frac{k-1}{(k-2)!}-\frac{k}{k!}=\frac{(k - 2) k^2}{k!}>0 \implies \bigg|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}\bigg|=\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}$$

Then as @Archis said use telescoping sum.