How to prove that $\frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<(1+\frac{1}{m})^{n+1}\frac{m^{n+1}}{n+1} $?

Question

The first part of the problem is:

Prove that for all integers $n \ge 1$ and real numbers $t>1$, $$ (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1)$$

I have done the first part by induction on $n$ for any real $t>1$.

However, I don't know how to do the second part, which is:

Use this to prove that if $m$ and $n$ are positive integers,

$$ \frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1} $$

I factorized $t^{n+1}-1$ into $(t-1)(1+t+t^2+\dots+t^n)$ and cancelled $t-1$ to obtain $(n+1)t^n>1+t+t^2+\dots+t^n>n+1$. And have no idea what to do next. This is the only way I could think of in order to get a "sum", but couldn't see any relation between the two sum (if there are any...).

The last part of the question is:

Find $$ \lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}}$$

I think I know how to do the last part. It should be divide the inequality by $m^{n+1}$, then apply Squeeze Theorem. I believe the answer is $\frac{1}{n+1}$.

Answer 1

We only need to prove the second part of question.

For the left inequality, let's apply induction on $m$.

Base case: $\frac{1}{n+1}<1$, trivial.

Induction case: Note that $1^n+\cdots+m^n>\frac{(m-1)^{n+1}}{n+1}+m^n$ by induction hypothesis. By first part, $(1+\frac{1}{m-1})^{n+1}-1<(n+1)(1+\frac{1}{m-1})^{n}\frac{1}{m-1}$, which is equivalent to $m^{n+1}-(m-1)^{n+1}<(n+1)m^{n}$ or $(m-1)^{n+1}>m^{n+1}-(n+1)m^{n}$. Applying it, we get $$\frac{(m-1)^{n+1}}{n+1}+m^n>m^n+\frac{m^{n+1}-(n+1)m^{n}}{n+1}=\frac{m^{n+1}}{n+1}$$ therefore induction case is also proved.

For the right inequality, it's simpler.

From the first part, $\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1}>\frac{m^{n+1}}{n+1}(\frac{n+1}{m}+1)=m^n+\frac{m^{n+1}}{n+1}$. Now apply induction on $m$.

For the base case, it's $1<\frac{2^{n+1}}{n+1}$ which is not hard.

For the induction case, from the induction hypothesis and above inequality,$$1^n+\cdots+m^n<\Big(1+\frac{1}{m-1}\Big)^{n+1}\frac{(m-1)^{n+1}}{n+1}+m^n=\frac{m^{n+1}}{n+1}+m^n<\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1}$$

Answer 2

Hint:

Note that $x^n$ is convex when $1 \le n$.
Thus you can profitably demonstrate the inequality through a Riemann sum of the integral of $x^n$ (histogram below and above the continuous curve).

Answer 3

Let $S=1^n+2^n+3^n+\cdots+m^n$.

From a quick graph sketch it is clear that: $$\begin{align} \int_0^m x^n dx \;\;&<\qquad \qquad \quad S &&< \int_0^m (x+1)^n dx\\ \left[\frac {x^{n+1}}{n+1}\right]_0^m\;\;&<\qquad \qquad \quad S &&<\;\;\left[\frac {(x+1)^{n+1}}{n+1}\right]_0^m\\ \frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\frac {(m+1)^{n+1}}{n+1}-\frac 1{n+1}\\ \frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\left[\left(1+\frac 1m\right)^{m+1}-1\right]\frac {m^{n+1}}{n+1}<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\ \frac {m^{n+1}}{n+1}\;\;&<\qquad1^n+2^n+3^n+\cdots+m^n &&<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\ \end{align}$$