Show that $2\leqslant\sum_{n=0}^\infty \frac{1}{n!}\leqslant3 $ 
Show that $$2\leqslant\sum_{n=0}^\infty \frac{1}{n!}\leqslant3 $$

I know that $$\sum_{n=1}^\infty \frac{n}{(n+1)!}=1 $$
Since $$ \frac{n}{(n+1)!} = \frac{1}{n!}- \frac{1}{(n+1)!} $$ 
I'm thinking that this equation might be helpful on deducing the inequality above, however I'm not sure on how to do it. 
 A: You had the right idea. Obviously
$$\sum_{n\geq 0}\frac{1}{n!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\ldots > 1+1 = 2, $$
but $\sum_{n\geq 1}\frac{n}{(n+1)!}=1$ ensures 
$$\sum_{n\geq 0}\frac{1}{n!}=2+\sum_{n\geq 1}\frac{1}{(n+1)!}<2+\sum_{n\geq 1}\frac{n}{(n+1)!}=3.$$
A: Simply note that since by esponential series
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
for $x=1$
$$2<e=\sum_{n=0}^\infty \frac{1}{n!}<3$$
For a direct prove note that
$$\sum_{n=0}^3 \frac{1}{n!}>2$$
and note that since $k!\ge 2^{k-1}$ (simple to be proved by induction)
$$\sum_{k=0}^n \frac1{k!}=1+\sum_{k=1}^n \frac1{k!}\le 1+\sum_{k=1}^n \frac1{2^{k-1}}=1+\sum_{k=0}^{n-1} \frac1{2^k}$$
and note also that by geometric series
$$\sum_{k=0}^{n-1} \frac1{2^k}=\frac{1-\frac1{2^n}}{1-\frac12}=2\left(1-\frac1{2^n}\right)<2$$
A: You need to find lower and upper bounds for this sum. The lower bound $2$ is simply to obtain by looking at the first two terms in the sum. The upper bound can be obtained by the following estimation
$$1+1/1!+\sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}\leq 1 +1 +\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)} $$
$$= 2 + \sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}=2+\sum_{n=1}^{\infty}\left[\dfrac{1}{n}-\dfrac{1}{n+1}\right]$$
The remaining sum can be calculated by telescoping.
$$\sum_{n=1}^{N}\left[\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\left[\dfrac{1}{1}-\dfrac{1}{2}\right]+\left[\dfrac{1}{2}-\dfrac{1}{3}\right]+\ldots+\left[\dfrac{1}{N}-\dfrac{1}{N+1}\right]$$
$$=\dfrac{1}{1}-\dfrac{1}{N+1} \to 1 \,(-0) \text{ for } N \to \infty$$
The $(-0)$ just is there to remind that the limit approaches $1$ from below.
