# Thread: Series proof

1. ## Series proof

Let p and n be real. Now show that
$
n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p \\
$

using the identity:
$
b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \dotsb + ba^{n-2} + a^{n-1}), n \in \mathbb{N}
$

I have been trying at this one for hours, it is driving me nuts. Can someone point me in the right direction ?

2. Originally Posted by hjortur
Let p and n be real. Now show that
$
n^p \le \frac{(n+1)^{p+1} - n^{p+1}}{p+1} \le (n+1)^p \\
$

using the identity:
$
b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \dotsb + ba^{n-2} + a^{n-1}), n \in \mathbb{N}
$

I have been trying at this one for hours, it is driving me nuts. Can someone point me in the right direction ?
This only works for $n,p\in\mathbb{N}$.

Notice that

$(n+1)^{p+1} - n^{p+1}=[(n+1)-n][(n+1)^p+(n+1)^{p-1}n+...+(n+1)n^{p-1}+n^p]$

$=(n+1)^p+(n+1)^{p-1}n+...+(n+1)n^{p-1}+n^p$.

Now notice that each term in this series is represented by

$(n+1)^{p-i}n^i$,

and that there are a total of $p+1$ terms therein. Notice also that for each term, we have

$n^p\leq(n+1)^{p-i}n^i\leq(n+1)^p$,

which means that for the entire series we have

$(p+1)n^p\leq\sum_{i=0}^p(n+1)^{p-i}n^i\leq(p+1)(n+1)^p$.

Substituting, we have

$(p+1)n^p\leq(n+1)^{p+1} - n^{p+1}\leq(p+1)(n+1)^p$.

Then we divide by $p+1$:

$n^p\leq\frac{(n+1)^{p+1} - n^{p+1}}{p+1}\leq(n+1)^p$. $\blacksquare$

3. Yeah, sorry for that, I was mistranslating the word "real" .

Also there was a typo (i am pretty new at this latex stuff) it was not supposed to be "less-than-or-equal" but "less-than". It is fixed above.

Notice also that for each term, we have
$
n^p\leq(n+1)^{p-i}n^i\leq(n+1)^p
$
so this one is supposed to be:
$
n^p<(n+1)^{p-i}n^i<(n+1)^p
$

wich does not hold for i=0.

Sorry for that.

4. Originally Posted by hjortur
Yeah, sorry for that, I was mistranslating the word "real" .

Also there was a typo (i am pretty new at this latex stuff) it was not supposed to be "less-than-or-equal" but "less-than". It is fixed above.

so this one is supposed to be:
$
n^p<(n+1)^{p-i}n^i<(n+1)^p
$

wich does not hold for i=0.

Sorry for that.
The strict inequalities hold as long as $i\neq 0$ and $i\neq p$, which is satisfied for at least one term in the series so long as $p\geq 2$.