$\sum _{k=1}^{n-1}k^p \lt \frac {n^{p+1}} {p+1} \lt \sum _{k=-1}^{n}k^p$

Question

I'm going through a proof of the theorem that says $\int_0^bx^pdx = \frac {b^{p+1}}{p+1}$, and it begins with the inequality. $\sum _{k=1}^{n-1}k^p \lt \frac {n^{p+1}} {p+1} \lt \sum _{k=-1}^{n}k^p$ What I'm having trouble understanding where this middle term came from.

Answer 1

Since this inequality is being used to prove an integral, perhaps a non-calculus proof would be best.

At the end of this answer is a simple inductive proof of Bernoulli's Inequality:

For non-negative integer $n$ and $x\ge-1$, $$ 1+nx\le(1+x)^n\tag{1} $$ Applying $(1)$ $$ \begin{align} 1-\frac{p+1}{k}&\le\left(1-\frac1k\right)^{p+1}\\ k^{p+1}-(p+1)k^p&\le(k-1)^{p+1}\\ k^{p+1}-(k-1)^{p+1}&\le(p+1)k^p\tag{2} \end{align} $$ Applying $(1)$ $$ \begin{align} 1+\frac{p+1}{k-1}&\le\left(1+\frac1{k-1}\right)^{p+1}\\ \frac{p+1}{k-1}&\le\left(1+\frac1{k-1}\right)^{p+1}-1\\ (p+1)(k-1)^p&\le k^{p+1}-(k-1)^{p+1}\tag{3} \end{align} $$ Combining $(2)$ and $(3)$, $$ (p+1)(k-1)^p\le k^{p+1}-(k-1)^{p+1}\le(p+1)k^p\tag{4} $$ Summing $(4)$ from $1$ to $n$ and dividing by $p+1$ yields $$ \sum_{k=0}^{n-1}k^p\le\frac{n^{p+1}}{p+1}\le\sum_{k=1}^nk^p\tag{5} $$ Note that for $p=0$, we have equalty in $(5)$. I've added a strict case to the proof of Bernoulli that, when replacing $(1)$ for $n\ge2$, yields that for $p\ge1$, we have strict inequality: $$ \sum_{k=0}^{n-1}k^p\lt\frac{n^{p+1}}{p+1}\lt\sum_{k=1}^nk^p\tag{6} $$

Answer 2

I didn't realize I was going to get something slightly different of what you're being given when I started to write this, but I leave you an idea to prove that $$\int_0^a x^pdx=\frac{a^{p+1}}{p+1}$$ The idea is the following:

PROP Let $$A(n)=\sum_{k=1}^n k^p$$

Then $A(n)=\frac{n^{p+1}}{p+1}+P_p(n)$ where $\operatorname{deg}P\leq p$

PROFF By induction on $p$.

For $p=1$, we obtain $$\sum\limits_{k = 1}^n k = {{{n^2}} \over 2} + {n \over 2}$$ Assume true for $m\leq p-1$ and consider $p$. We exploit the fact that $${(k + 1)^{p+1}} - {k^{p+1}} = \sum\limits_{m = 0}^{p} {p+1\choose m}{{k^m}} $$

We then have, after summing from $k=1$ to $p$ that $${(n + 1)^{p + 1}} - 1 = \sum\limits_{m = 0}^p {\left( \matrix{ p + 1 \cr m \cr} \right)} \sum\limits_{k = 1}^n {{k^m}} $$

or

$${(n + 1)^{p + 1}} - 1 = \sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} \sum\limits_{k = 1}^n {{k^m}} + \left( {p + 1} \right)\sum\limits_{k = 1}^n {{k^p}} $$ Now, our inductive hypothesis is that

$$\sum\limits_{k = 1}^n {{k^m}} = {{{n^{m + 1}}} \over {m + 1}} + {P_m}\left( n \right)$$

This gives us that

$${{{{(n + 1)}^{p + 1}}} \over {p + 1}} - {1 \over {p + 1}} - {1 \over {p + 1}}\sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} \left( {{{{n^{m + 1}}} \over {m + 1}} + {P_m}\left( n \right)} \right) = \sum\limits_{k = 1}^n {{k^p}} $$

Now, let's look at that big mess above. Observe that by the binomial theorem, we can write $${{{{(n + 1)}^{p + 1}}} \over {p + 1}} = {{{n^{p + 1}}} \over {p + 1}} + {Q_p}\left( n \right)$$ where the degree of $Q_p$ is $p$. Note also that

$$\sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} \left( {{{{n^{m + 1}}} \over {m + 1}} + {P_m}\left( n \right)} \right) = \sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} \left( {{{{n^{m + 1}}} \over {m + 1}}} \right) + \sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} {P_m}\left( n \right)$$

The first term on the right is a polynomial of degree $p$. On the other hand, the sum of all the polynomials on the left is of degree $\leq p$, since it's a sum of polynomials of degree at most $p$. All in all, we may write $${{{{(n + 1)}^{p + 1}}} \over {p + 1}} - {1 \over {p + 1}} - {1 \over {p + 1}}\sum\limits_{m = 0}^{p - 1} {\left( \matrix{ p + 1 \cr m \cr} \right)} \left( {{{{n^{m + 1}}} \over {m + 1}} + {P_m}\left( n \right)} \right) = {{{n^{p + 1}}} \over {p + 1}} + {W_p}\left( n \right)$$ where $W$ has degree at most $p$. Thus $$\sum\limits_{k = 1}^n {{k^p}} = {{{n^{p + 1}}} \over {p + 1}} + {W_p}\left( n \right)$$ and the proposition is proven.

Now, we look at our integral approximation, which is $${1 \over n}\sum\limits_{k = 1}^n {{{\left( {{{ak} \over n}} \right)}^p}} $$

Because of the above, we have that $${1 \over n}\sum\limits_{k = 1}^n {{{\left( {{{ak} \over n}} \right)}^p}} = {{{a^{p + 1}}} \over {p + 1}} + {a^{p + 1}}{{{W_p}\left( n \right)} \over {{n^{p + 1}}}}$$

Since $W_p(n)$ has degree at most $p$, it follows that $${{{W_p}\left( n \right)} \over {{n^{p + 1}}}}\to 0$$

when $n\to \infty$, so we conclude that

$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{{\left( {{{ak} \over n}} \right)}^p}} = {{{a^{p + 1}}} \over {p + 1}}$$

Answer 3

We use the fact that $$a^p-b^p=(a-b)\sum_{k=0}^{p-1}a^kb^{p-k-1}$$ Because of that, we have $$\tag 1 n^p<\frac{(n+1)^{p+1}-n^{p+1}}{p+1}<(n+1)^p$$

Indeed, by the above $${\left( {n + 1} \right)^{p + 1}} - {n^{p + 1}} = \sum\limits_{k = 0}^p {{n^k}} {\left( {n + 1} \right)^{p - k}}$$

that is $${\left( {n + 1} \right)^{p + 1}} - {n^{p + 1}} = \sum\limits_{k = 0}^p {{{\left( {{n \over {n + 1}}} \right)}^k}} {\left( {n + 1} \right)^p}$$

but for any choice of $n$ $${n \over {n + 1}} < 1$$ so that$${\left( {n + 1} \right)^{p + 1}} - {n^{p + 1}} = \sum\limits_{k = 0}^p {{{\left( {{n \over {n + 1}}} \right)}^k}} {\left( {n + 1} \right)^p} < \sum\limits_{k = 0}^p {{{\left( {n + 1} \right)}^p} = \left( {p + 1} \right)} {\left( {n + 1} \right)^p}$$ and one side is done. But similarily, $${{n + 1} \over n} > 1$$ for any choice of $n$ so $$\left( {p + 1} \right){n^p} = \sum\limits_{k = 0}^p {{n^p}} < \sum\limits_{k = 0}^p {{{\left( {{{n + 1} \over n}} \right)}^{p - k}}{n^p}} = \sum\limits_{k = 0}^p {{{\left( {n + 1} \right)}^{p - k}}{n^k}} = {\left( {n + 1} \right)^{p + 1}} - {n^{p + 1}}$$ and we get what we wanted.

What you want follows from summing telescopically the inequalities in $(1)$.

Answer 4

For $p>0$ this can be proved easily,

$\sum _{k=1}^{n-1}k^p < \frac {n^{p+1}} {p+1} < \sum _{k=-1}^{n}k^p$

Suppose this is true for $n=m$

Then for $n=m+1$ we have,

$\sum _{k=1}^{m}k^p = \sum _{k=1}^{m-1}k^p+m^p$

If we can show that $m^p<\displaystyle \frac{(m+1)^{p+1}-m^{p+1}}{p+1} $ then we are done.

This can easily be proved using mean value theorem,

Consider the function $f(x)=x^{p+1}$,

Applying mean value theorem,

$\exists c\in(m,m+1)$ such that ,

$\displaystyle \frac{(m+1)^{p+1}-m^{p+1}}{m+1-m}=f'(c)$

$\displaystyle \frac{(m+1)^{p+1}-m^{p+1}}{m+1-m}=(p+1)c^p>(p+1)m^p$

$\displaystyle \Rightarrow \frac{(m+1)^{p+1}-m^{p+1}}{p+1}>m^p$

The right side can also be proved similarly.