Question on Summation and Geometric Progression Let $S_k$, $k=1,2,3,....,100$, denote the sum of the infinite geometric progression series whose first term is $\displaystyle\frac{k-1}{k!}$ and the common ratio is $\displaystyle\frac1k$. Then  find the value of $$\frac{100^2}{100!}+\displaystyle\sum_{k=1}^{100}|(k^2-3k+1)S_k|$$
I could proceed as follows:
The given first term of the GP and the common ration gives the sum as $\displaystyle\frac{1}{(k-1)!}$. However, I am unable to solve the series after putting $S_k$ in the expression.

Writing $k^2-3k+1 = (k^2-3k+2)-1 = (k-1)(k-2)-1$, we see that the term of the sum for $k \geqslant 3$ is
$$(k-1)(k-2)S_k - S_k = S_{k-2} - S_k,$$
so overall we get a telescoping sum
$$\begin{align}
\sum_{k=1}^{100} \lvert (k^2-3k+1)S_k\rvert &= S_1 + S_2 + \sum_{k=3}^{100} S_{k-2}-S_k\\
&= 2(S_1 + S_2) - S_{99} -S_{100}\\
&= 4 - \frac{1}{98!} - \frac{1}{99!}\\
&= 4 - \frac{100}{99!}\\
&= 4 - \frac{100^2}{100!}.
\end{align}$$
