Cool summation

For $k \neq 1$ we get that

$S{_{k}}= \frac{k-1} { k!} \times \frac{1}{1 - \frac{1}{k} } = \frac{1}{(k-1)!}$

For $k = 1$ , notice that the first term is 0, and hence the sum is $S_1 = 0$ . The reason why we do not get $S_1 = \frac{1}{ ( 1 - 1)! }$ is because we had to cancel out by $k-1 = 0$ in the algebraic manipulation, which would not be allowed otherwise.

It is also fairly obvious that the question is an exercise in telescoping. So for now, we ignore the modulus in the given expression and work with:

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

Thus, the general term is:

$\frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}$

We can now telescope this quite easily. But it is now that the modulus comes into play:

$k^2-3k+1$ is negative for $k=1$ and $k=2$ . This can be taken into account by telescoping from $k=3$ to $k=100$ and adding the terms that remain.

As such, splitting it out, we obtain

$\frac{100^2}{100!} + | 0| + \left| ( 2^2 - 3 \times 2 + 1) \times \frac{1}{1!} \right| + \left| \frac{ 3 -1 } { (3-2) ! } - \frac{ 100} { (100-1)! } \right|$

This evaluates out to 3.