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$(U_n)$ and $(V_n)$ are two sequences such that :

$U_n = \frac{x^{p+1}}{(p+1)!} +\frac{x^{p+2}}{(p+2)!} +...+\frac{x^{p+n}}{(p+n)!}$

$V_n = 1+\frac{x}{p+1}+\frac{x^2}{(p+1)^2}+...+\frac{x^{n-1}}{(p+1)^{n-1}}$

where $~~p \in \mathbb{N}~~~, ~~x\in \mathbb{R}~~~~~~$ and $~~~~ 0\le x < p+1$ .

prove that : $~~~~~~ U_{n} \le ~\frac{x^{p+1}}{(p+1)!} . V_{n} $

by developing $~~\frac{x^{p+1}}{(p+1)!} . V_{n}~~$ we get :

$\frac{x^{p+1}}{(p+1)!} . V_{n} = \frac{x^{p+1}}{(p+1)!} +\frac{x^{p+2}}{(p+1)!(p+1)} +...+\frac{x^{p+n}}{(p+1)!(p+1)^{n-1}}$

so in order to prove the inequality above , we just have to prove that :

$(p+n)!\le (p+1)!(p+1)^{n-1} $

and that's where I'm blocked, because I don't know how to prove the inequality above. Any tips ?

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    What you have to prove is that $(p+k)! \ge (p+1)!(p+1)^{k-1}$ for all $k \ge 1$. Hint; $(p+k)! = (p+k)(\cdots)(p+2)(p+1)!$. – 2017-01-27
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    My first choice is mathematical induction – 2017-01-27

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As Open Ball has already pointed out, you have the inequality reversed. Since it involves the denominators, the larger denominator gives the smaller ratio.

Now, just express $$\frac{(p+n)!} {(p+1)!(p+1)^{n-1}}= \frac{(p + 2)(p + 3)...(p+n)}{(p+1)(p+1)...(p+1)} > 1$$