Let $S_k$, $k = 1, 2, ....., 100$, 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 $\frac{100^2}{100!} + \sum_{k=1}^{100} |(k^2 - 3k + 1)S_k|$ is

Let Sₖ, k = 1, 2, ....., 100, denote the sum of the infinite geometric series whose first term is (k-1)/k! and the common ratio is 1/k. Then the value of 100²/100! + ∑ₖ₌₁¹⁰⁰ |(k² - 3k + 1)Sₖ| is

$a = \dfrac{k-1}{k!}, \qquad r = \dfrac{1}{k}$

$S_k = \dfrac{k-1}{k!} + \dfrac{k-1}{k!}\cdot\dfrac{1}{k} + \dfrac{k-1}{k!}\cdot\dfrac{1}{k^2} + \cdots$

$\text{To find: } \dfrac{100^2}{100!} + \sum_{k=1}^{100} \left| (k^2 - 3k + 1)\,S_k \right|$

$\text{Plan: } (1)\ \text{closed form of } S_k, \quad (2)\ \text{simplify the general term}$

$(3)\ \text{telescope the sum, minding the modulus}$

$S_k = \dfrac{a}{1-r}$

$= \dfrac{\dfrac{k-1}{k!}}{1-\dfrac{1}{k}} = \dfrac{\dfrac{k-1}{k!}}{\dfrac{k-1}{k}}$

$S_k = \dfrac{k}{k!} = \dfrac{1}{(k-1)!}$

$(k^2-3k+1)\,S_k = \dfrac{k^2-3k+1}{(k-1)!}$

$k^2-3k+1 = (k-1)(k-2) - 1$

$\dfrac{k^2-3k+1}{(k-1)!} = \dfrac{(k-1)(k-2)}{(k-1)!} - \dfrac{1}{(k-1)!}$

$= \dfrac{1}{(k-3)!} - \dfrac{1}{(k-1)!}$

$k^2-3k+1 > 0 \quad \text{for } k \ge 3$

$k=1,2: \quad (k^2-3k+1)S_k = -1$

$\sum_{k=1}^{100} |(k^2-3k+1)S_k| = 1 + 1 + \sum_{k=3}^{100} (k^2-3k+1)S_k$

$= 2 + \sum_{k=3}^{100} \left( \dfrac{1}{(k-3)!} - \dfrac{1}{(k-1)!} \right)$

$\sum_{k=3}^{100} \dfrac{1}{(k-3)!} = \dfrac{1}{0!} + \dfrac{1}{1!} + \cdots + \dfrac{1}{97!}$

$\sum_{k=3}^{100} \dfrac{1}{(k-1)!} = \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{99!}$

$\sum_{k=3}^{100} \left( \dfrac{1}{(k-3)!} - \dfrac{1}{(k-1)!} \right) = 2 - \dfrac{1}{98!} - \dfrac{1}{99!}$

$\sum_{k=1}^{100} |(k^2-3k+1)S_k| = 4 - \dfrac{1}{98!} - \dfrac{1}{99!}$

$\dfrac{100^2}{100!} = \dfrac{10000}{100!}$

$\dfrac{1}{98!} + \dfrac{1}{99!} = \dfrac{9900}{100!} + \dfrac{100}{100!} = \dfrac{10000}{100!}$

$\text{Answer: } 4 - \dfrac{10000}{100!} + \dfrac{10000}{100!} = 4$