Which of the following statements describe the sum $S_n = {\sum\limits_{k=2}^{n} \frac{k^2-2}{k!}}$ correctly?

Question

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.

Answer 1

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.

Answer 2

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.

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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 $n\to \infty$?