Sum of an infinite geometric sum

Firstly,it is clear that, $f(1) = 0$ .

Next consider $f(t)$ ,
$f(t) = \frac{t-1}{t!} \times \frac{1}{(1-\frac{1}{t})}$
$\Rightarrow f(t) = \frac{(t-1)t}{t!(t-1)}$
For $t \neq 1$ ,
$f(t) = \frac{1}{(t-1)!}$

Now consider the general term of summation,
$= (r^2 -3r +1)f(r)$
$= (r^2 -2r +1 - r)\frac{1}{(r-1)!}$
$= ((r-1)^2 -r)\frac{1}{(r-1)!}$
$= \frac{(r-1)!}{(r-2)!} - \frac{r}{(r-1)!}$
Clearly, the above expression forms a telescopic series and summing it over $r=3$ to $r=100$ ,we get,
$\sum_{r=3}^{100} |(r^2 - 3r +1)f(r)| = \frac{2}{1!} - \frac{100}{99!}$

Therefore,
$S(100) = |(1- 3 + 1)f(1)| + |(4 - 6 +1)f(2)| + \sum_{r=3}^{100} |(r^2 - 3r +1)f(r)|$
$S(100) = 0 + 1 + 2 - \frac{100}{99!}$

Thus the required value,
$= \frac{100^2}{100!} + S(100)$
$= \frac{100}{99!} + 3 - \frac{100}{99!}$ = $\boxed{3}$

ok, so looking at the summation first, what better way to do a summation than summing everything? There are a lot of better ways, but anyway, to get a feel of what we're looking at, let's sum the first couple terms after setting a definition first so it's easier for me to write:

$h\left( x \right) =|({ x }^{ 2 }-3x+1)(f\left( x \right) )|$

Okay:

$r=1, f(r)=0/1+0/1+0/1+...=0, h(r)=|(1-3+1)(0)|=0$

$r=2, f(r)=1/2+1/4+1/8+...=1, h(r)=|(4-6+1)(1)|=1$

$r=3, f(r)=1/2, h(r)=|(9-9+1)(1/2)|=1/2$

Okay, it should be fairly clear that after $r=2$ , then $r^2-3r+1$ is always going to be positive. Intuitively this is because after $r=3$ then $r^2$ is going to be bigger than $3r$ . Okay so now that we have done that, let's simplify $f(x)$ using our handy-dandy infinite sum formula:

$a+\frac{a}{r}+\frac{a}{r^2}+...=\frac{a}{1-r}$

And this is only when r<1 but we don't have to worry about that because $\frac{1}{t}<1$ is always true if t is bigger than 1. If t=1, then we have f(t)=0 as I found up there. Now let's just plug in:

$f(r)=\frac{\frac{t-1}{t!}}{1-\frac{1}{t}}=\frac{1}{(t-1)!}$

which we get after simplifying. Now let's look at what we have to find simplified:

\sum_{r=1}^{100} {|(r^2-3r+1)(\frac{1}{(r-1)!})|})

That $r^2-3r+1$ looks eerie. It looks as if there's no point to it being there. Let's try to incorporate that into helping us; after all, it's not just there for show. We know it can't factor. The 1 would have to be a 2 for it to factor nicely. But, wait a minute, if it was a 2, then it would factor as $r^2-3r+2=(r-2)(r-1)$ but we have a $(r-1)!$ in the denominator! We can cancel those! Let's try to do this by making $r^2-3r+1=(r-2)(r-1)-1$ :

$h(r)=|\frac{(r-2)(r-1)-1}{(r-1)!}|=|\frac{(r-2)}{(r-2)!}-\frac{1}{(r-1)!}|=\frac{1}{(r-3)!}-\frac{1}{(r-1)!}$

You notice I took away the absolute signs. This is because we don't have to worry about it ever being negative because, sinc r-3 is smaller than r-1, then $1/{(r-3)!}$ is always bigger than $1/(r-1)!$ . Now, you also notice this is a telescoping sum! So THAT was the point of the quadratic there! However we only run into problems when r is less than 3 because we have negative factorials. Don't worry; we have already calculated when r=1 and r=2. Therefore we need to find:

$0+1+\sum_{r=3}^{100} {\frac{1}{(r-3)!}-\frac{1}{(r-1)!}}$

Okay so, finally the telescoping summation cancels everything but the last 2 terms and the first 2 terms:

$\sum_{r=3}^{100} {h(r)}=(1-{1/2})+(1-{1/6})+({1/2}-{1/24})+({1/6}-{1/120})+...+({1/96!}-1/98!)+({1/97!}-1/99!)=2-1/98!-1/99!$

Finally, we have to simplify:

$\frac{100^2}{100!}+0+1+2-\frac{1}{98!}-\frac{1}{99!}=3-\frac{1}{98!}-\frac{1}{99!}+\frac{100}{99!}=3-\frac{1}{98!}+\frac{99}{99!}=3-\frac{1}{98!}+\frac{1}{98!}=\boxed{3}$