How to derive these inequalities?

Question

I can derive the inequalities $$ n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p $$ for any positive integers $p$ and $n$. These actually follow from the identity $$b^p - a^p = (b-a)(b^{p-1} + b^{p-2}a + b^{p-3}a^2 + \ldots + b^2 a^{p-3} + b a^{p-2} + a^{p-1} ). $$

Now how to derive from the above the following inequalities? $$ \sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^{n} k^p, $$ where $p$ and $n$ are positive integers.

Answer 1

Hint: rewrite the starting inequality with $n$ replaced by $k$. Sum over $k$. Note one of the sums is telescopic.

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Edit:

For positive integers $n$ and $p$,

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

So for positive integers $k$ and $p$,

$$k^p < \frac{(k+1)^{p+1} - k^{p+1}}{p+1}\\ \implies \sum_{k=1}^{n-1} k^p < \sum_{k=1}^{n-1} \left(\frac{(k+1)^{p+1} - k^{p+1}}{p+1}\right)=\frac{n^{p+1} - 1^{p+1}}{p+1}<\frac{n^{p+1}}{p+1}\\ \implies \sum_{k=1}^{n-1} k^p <\frac{n^{p+1}}{p+1}.$$

That takes care of the left side of the inequality you are trying to prove. The right side is completely analogous.

Answer 2

We estimate this by integrals. See the middle term is $\displaystyle \int_{0}^{n}x^{p}\,\mathrm{d}x$. Now right side is the total area under $[0,n]$ for $x^p$ due overestimation by rectangles and similarly underestimating by rectangles, i.e counting one rectangle less, we get the left inequality. It is clear if you have a general knowledge of integrals and draw the graph of $x^p$.

One other way would be through induction. See the first inequality you posed, gives $$\begin{align} & n^p+\frac{n^{p+1}}{p+1}<\frac{(n+1)^{p+1}}{p+1}<(n+1)^p+\frac{n^{p+1}}{p+1}\\ & \implies \sum_{i=1}^{n-1}i^p+n^p<n^p+\frac{n^{p+1}}{p+1}<\frac{(n+1)^{p+1}}{p+1}<(n+1)^p+\frac{n^{p+1}}{p+1}<\sum_{i=1}^{n}i^p+(n+1)^p\\ \end{align}$$ This completes the induction.