Any simpler form for $ \frac{\sum_{k=2}^{n-2}{k\left(\sum_{i=0}^{k}\frac{(-1)^i}{i!}\right)}}{n\sum_{i=0}^{n}\frac{(-1)^i}{i!}}$ Is there any simpler form for the following expression:
$$
\frac{\sum_{k=2}^{n-2}{k\left(\sum_{i=0}^{k}\frac{(-1)^i}{i!}\right)}}{n\sum_{i=0}^{n}\frac{(-1)^i}{i!}}$$
Because I have to compute this expression for $n=100$, and without using any computing packages, doing this would be really tedious and painful.
A few days back, I posted a question: Any simpler expression for$\frac{\sum_{k=2}^{n-2}{k\big(\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}\big)}}{n\sum_{i=0}^{n}\frac{(-1)^i}{i!}}$ . While the answers are amazing, when I was reading the answers, I was so confused and just realised that I actually typed in the wrong expression.
Any help would be highly appreciated!

Changing the order of summation and noting that the terms for $i=0, i=1$ cancel:
$$\sum^{n-2}_{k=2} \sum_{i=0}^k k \frac{(-1)^i}{i!}=\sum^{n-2}_{i=2} \sum_{k=i}^{n-2} k \frac{(-1)^i}{i!}$$
We have
$$\sum^{n-2}_{k=i} k = \frac{(n-2)(n-1)}{2} - \frac{(i-1)i}{2}$$
Plugging this into the previous expression we get
$$\frac{(n-2)(n-1)}{2}\sum_{i=2}^{n-2} \frac{(-1)^i}{i!} - \sum_{i=2}^{n-2} \frac{(i-1)i}{2}\frac{(-1)^i}{i!}$$
The second of these terms equals
$$\frac12\sum_{i=2}^{n-2} \frac{(-1)^i}{(i-2)!} = \frac12 \sum_{i=0}^{n-4} \frac{(-1)^i}{i!}$$
Now let us abbreviate $a_n = \sum_{i=0}^n \frac{(-1)^i}{i!}$. Note, $1\le a_n< e$ and $a_n\to e$ as $n\to\infty$. Then altogether we have simplified your expression to:
$$\frac{\sum^{n-2}_{k=2} \sum_{i=0}^k k \frac{(-1)^i}{i!}}{n \sum_{i=0}^n \frac{(-1)^i}{i!}} = \frac{1}{2 n a_n}\Big((n-2)(n-1)a_{n-2} - a_{n-4}\Big) =\frac{(n-2)(n-1)-1}{2n}+R_n$$
where $R_n$ is an error term given by
$$R_n=\frac{(n-2)(n-1)-1}{2n}\Big(\frac{a_{n-4}}{a_n}-1\Big)+ (-1)^{n-4} \frac{(n-2)(n-1)}{2n a_n} \Big(\frac1{(n-4)!}-\frac1{(n-3)!}\Big)$$
This tends to $0$ very quickly as $n\to\infty$. Estimate the first term using
$$\Big|\frac{a_{n-4}}{a_n}-1\Big|= \frac{|a_n-a_{n-4}|}{a_n} \le \frac{1}{n!}+\frac1{(n-1)!}+\frac1{(n-2)!}+\frac1{(n-3)!}\le \frac4{(n-3)!}$$
and the second term similarly so that we get
$$|R_n| < 10\frac{n^2}{(n-3)!}$$
For instance, $|R_{100}|< 10^{-147}$, which is comfortably beyond machine precision of any ordinary scientific calculator. 
So for $n=100$ your number is equal to
$$\frac{98\cdot 99-1}{200}+R_{100} = 48.505+R_{100}$$
