How to show $\frac{2}{2!}+\frac{7}{3!}+...+\frac{k^2-2}{k!}+...+\frac{9998}{100!}<3$

Question

Demonstrate the inequality$$\frac{2}{2!}+\frac{7}{3!}+...+\frac{k^2-2}{k!}+...+\frac{9998}{100!}<3$$

Attempt:

$$\sum^{100}_{n=2} \frac{n^2-2}{n!}$$

Note that $n^2-2 \leq n^2$, so we have:

$$\sum^{100}_{n=2} \frac{n^2-2}{n!} \leq \sum^{100 }_{n=2}\frac{n^2}{n !}=\sum^{100}_{n=2}\frac{n}{(n-1)!}$$

I would like to solve the problem in a more elementary way, without using a calculator. I thought this problem might have some application to the Bernoulli Inequality, but I have no idea.

Answer 1

$$\begin{align}\sum^{k}_{n=2} \frac{n^2-2}{n!}&=\sum^{k}_{n=2} \frac{n^2-n}{n!}+\frac{n-2}{n!}\\ \\ &=\sum^{k}_{n=2} \frac{1}{(n-2)!}-\frac{1}{n!}+\sum^{k}_{n=2} \frac{1}{(n-1)!}-\frac{1}{n!}\\ \\ &=1+1-\frac{1}{(k-1)!}-\frac{1}{k!}+1-\frac{1}{k!}\\ \\ &=3-\frac{1}{(k-1)!}-\frac{2}{k!}\\ \\ &<3\end{align}$$