Which of the following statements describe the sum $S_n = {\sum\limits_{k=2}^{n} \frac{k^2-2}{k!}}$ correctly? I have the sum:
$$S_n =  {\displaystyle \sum_{k=2}^{n} \dfrac{k^2-2}{k!}}$$
with $n \ge 2$. I am asked to choose which one of the following statements describe the sum accurately (only one option is right):
A. $S_n < 3$
B. $S_n > 3$
C. $S_n = e$
D. $S_n < 0$
E. $S_n = e - \dfrac{1}{2}$
I can see that answer D is clearly out, but I don't know how to choose the right answer from A, B, C, E. I tried completing the square into something like:
$$S_n =  {\displaystyle \sum_{k=2}^{n} \dfrac{k^2-2}{k!}} =$$
$$={\displaystyle \sum_{k=2}^{n} \dfrac{k^2+2k+1-2k-1-2}{k!}} $$
$$={\displaystyle \sum_{k=2}^{n} \dfrac{(k+1)^2-(2k+3)}{k!}}  $$
$$={\displaystyle \sum_{k=2}^{n} \dfrac{(k+1)^2} {k!}} - {\displaystyle \sum_{k=2}^{n} \dfrac{2k+3}{k!}}  $$
$$={\displaystyle \sum_{k=2}^{n} \dfrac{(k+1)^2}{k!}} - 2{\displaystyle \sum_{k=2}^{n} \dfrac{1}{(k-1)!}} - 3{\displaystyle \sum_{k=2}^{n} \dfrac{1}{k!}}$$
And I got stuck here. I don't think completing the square helps me all that much in this problem. 
 A: HINT:
Note that we can write the summand as 
$$\begin{align}
\frac{k^2-2}{k!}&=\frac{k}{(k-1)!}-\frac2{k!}\\\\
&=\frac{k-1+1}{(k-1)!}-\frac2{k!}\\\\
&=\frac{1}{(k-2)!}+\frac{1}{(k-1)!}-\frac2{k!}\\\\
&=\frac{1}{(k-2)!}-\frac{1}{(k-1)!}+2\left(\frac1{(k-1)!}-\frac1{k!}\right)
\end{align}$$
Now "telescope."  You will obtain a closed form expression from which it is easy to find an upper bound.

If you don't wish to proceed with finding an upper bound, simply realize that $S_n$ must be a rational number.  That eliminates two choices.  You already know that $S_n>0$.  So, that leaves choices $(A)$ and $(B)$.  Finally, note that $S_n$ is increasing monotonically.  What is the limit as $n\to \infty$?  
A: Approach $1$: The answer must be true for all $n \ge 2$, including $n = 2$. At $n = 2$, the sum is $1$. The only answer choice that would be valid for is A.
Approach $2$: If you wanted to reason more mathematically about it, first notice that each term is positive and the limit of the terms approaches $0$. We can then split the sum into the two parts $$\left( \sum_{k=2}^n \frac{k^2}{k!} \right) - \left( \sum_{k=2}^n \frac{2}{k!} \right)$$ The second part approaches $\frac{2}{0!} + \frac{2}{1!} - 2e = 4-2e$. The first part approaches $$\sum_{k=2}^n \frac{k}{(k-1)!} = \sum_{k=2}^n \left( \frac{1}{(x-2)!} + \frac{1}{(x-1)!} \right) = e + e-1 = 2e-1$$
Adding the two together, we get that the infinite sum approaches $(4-2e) + (2e-1) = 3$. Since the sum converges to $3$, and each term is positive, each partial sum must be less than $3$, which is answer choice A.
