Hi,
I must show that $\displaystyle \forall$ n from $\displaystyle \mathbb{N}$* $\displaystyle \forall$ p from ]0,1[:
$\displaystyle \frac{p}{(n+1)^{1-p}}\le (n+1)^p-n^p \ge \frac{p}{n^{1-p}}$ by using mean value theorem, can you help me please???
Hi,
I must show that $\displaystyle \forall$ n from $\displaystyle \mathbb{N}$* $\displaystyle \forall$ p from ]0,1[:
$\displaystyle \frac{p}{(n+1)^{1-p}}\le (n+1)^p-n^p \ge \frac{p}{n^{1-p}}$ by using mean value theorem, can you help me please???
I assume that you mean that $\displaystyle \mathbb{N}=\mathbb{Z}^+$.
Define $\displaystyle f(x)=x^p$ then $\displaystyle f'(x)=px^{p-1}=\frac{p}{x^{1-p}}$.
Then by mean value $\displaystyle \left( {\exists c \in (n,n + 1)} \right)\left[ {f'(c) = f(n + 1) - f(n)} \right]$ or $\displaystyle \frac{p}{c^{1-p}}=(n+1)^p-n^p$.
Note that $\displaystyle \frac{1}{n+1}<\frac{1}{c} <\frac{1}{n} $.
Do the algebra to finish.