Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then the problem is pretty straightforward.
However, I don't understand why the inequality of the hint is true? A simple example shows that the inequality doesn't hold! There are no further information given.
What am I missing? Is $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$ simply nonsense? Or did the professor forgot to mention some more assumptions?
 A: The hint was not correctly stated. The estimate suggested in my comment is a possible fix for the typo, but the following estimate is even better: $$(n+1)!y_m\leq \sum_{k=1}^m \frac 1{(n+2)^{k-1}}.$$ Taking limit on both sides as $m\rightarrow \infty $ yields $$(n+1)!\left(e-\sum_{k=0}^n\frac 1 {k!}\right)\leq \frac 1{1-\frac 1{n+2}}=\frac {n+2}{n+1}$$
$$\Rightarrow e-\sum_{k=0}^n\frac 1{k!}\leq\frac {n+2}{(n+1)^2}\cdot \frac 1{n!}<\frac 1{n\cdot n!}.$$ Since the left inequality trivially holds, this proves the result.
A: We have
$$\begin{align}y_m &= \frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\cdots(n+m)}\right] \\&<\frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots + \frac{1}{(n+1)^m}\right]\end{align}$$
Evaluating the finite geometric sum on the RHS it follows that
$$y_m  < \frac{1}{n!}\frac{\frac{1}{n+1}- \frac{1}{(n+1)^{m+1}}}{1 - \frac{1}{n+1}}= \frac{1}{nn!}\left[1 - \frac{1}{(n+1)^{m}} \right] \underset{ m \to \infty}\longrightarrow 
 \frac{1}{nn!}  $$
A: I think I got it right this time!!
$e-\sum_{0}^{n}\frac{1}{k!}<\frac{1}{n!n}$ is equivalent to $ e-e+\sum_{n+1}^{\infty }\frac{1}{k!}<\frac{1}{n!n}$ equivalent to
$\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+2)}+.....<\frac{1}{n!n}$ and hence to
$\frac{1}{n!}({\frac{1}{(n+1)}}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+....)<\frac{1}{n!n}$
But $\frac{1}{(n+1)(n+2)}<\frac{1}{(n+1)^{2}}, \,\,\frac{1}{(n+1)(n+2)(n+3)}<\frac{1}{(n+1)^{3}} $ etc so it suffices to show setting $a=\frac{1}{n+1}$
that $a+a^{2}+a^{3}+....\leq \frac{1}{n}\Leftrightarrow \frac{1}{1-a}-1\leq \frac{1}{n}$. But this is
$\frac{n+1}{n}-1=\frac{1}{n}$.   QED
