I've edited the post in order to add at the end what, I think, is the complete proof of these inequalities. I want to apologize by not having given a reply as early as those given by the users who gave me hints about this exercise. It took me a little of time but your answers were very helpful (:.

This is actually part (b) of the exercise. I was able to prove these inequalities by applying the Binomial Theorem, but I have no idea about how to do it using part (a), which I'll post here too.

(a) Let $b$ be a positive integer. Prove that: $$b^p - a^p = (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})$$

(b) Let $p$ and $n$ denote positive integers. Use part (a) to show that

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

(a) Proof:

$$\begin{split}b^p - a^p &= (b-a)(b^{p-1}+ b^{p-2}a+b^{p-3}a^{2}+...+ba^{p-2}+a^{p-1})\\ &= (b-a)\sum_{k=0}^{p-1}\big[b^{p-(k+1)}a^k\big]\\ &=\sum_{k=0}^{p-1}\big[b^{p-(k+1)+1}a^k- b^{p-(k+1)}a^{k+1}\big]\quad \Leftarrow \small{\text{Distribute }(b-a).}\\ &=\sum_{k=0}^{p-1}\big[b^{p-k}a^k- b^{p-(k+1)}a^{k+1}\big]\\ &=b^{p-0}a^0- b^{p-[(p-1)+1]}a^{(p-1)+1}\quad\Leftarrow \small{\text{Apply the telescoping property for sums.}}\\ &= b^p - a^p\end{split}$$

(b) This is my attempt:

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

By multiplying by $p+1$ we have

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

which can also be written as

$$(p+1)n^p < [(n+1)-n]\sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p\\ (p+1)n^p < \sum_{k=0}^{p}(n+1)^{p-(k+1)}n^{k} < (p+1)(n+1)^p$$

By dividing the inequalities by $(n+1)^p$ we get

$$(p+1)\left(\frac{n}{n+1}\right)^p < \sum_{k=0}^{p}\frac{n^k}{(n+1)^{k+1}} < (p+1)$$

Here I'm stuck. I guess the last step wasn't necessary.


This is my last attempt. Hopefully it's not flawed.

We will prove each bound separately.

To prove: $(p+1)n^p < (n+1)^{p+1}-n^{p+1}$

Proof (direct):

Let the numbers $a$ and $b$ be defined as

$$a = n \quad \text{and}\quad b = n+1\qquad \text{For }n \in \mathbb{N}.$$

Then we have

$$a < b$$

And it follows

$$\frac{a^p}{a^k} < \frac{b^p}{b^k}\quad \text{For } p\ \text{and }k\in\mathbb{N}, p \neq k. \qquad (1)$$

By multiplying both sides by $a^k$ we get

$$a^p < b^{p-k}a^k$$

By taking the sum of both sides we have

$$\sum_{k=0}^pa^p < \sum_{k=0}^pb^{p-k}a^k$$

On the LHS $a^p$ is summed $p+1$ times (from $0$ to $p$). So it can also be written as

$$(p+1)a^p < \sum_{k=0}^pb^{p-k}a^k$$

Since $b-a = 1$, let multiply the RHS by $b-a$

$$(p+1)a^p < (b-a)\sum_{k=0}^pb^{p-k}a^k$$

By distributing $b-a$ inside the sum we have

$$(p+1)a^p <\sum_{k=0}^p[b^{p-(k-1)}a^k - b^{p-k}a^{k+1}]$$

And by applying the telescoping property for sums we get

$$\begin{align*}(p+1)a^p &< b^{p-(0-1)}a^0 - b^{p-p}a^{p+1}\\ (p+1)a^p &< b^{p+1} - a^{p+1}\\ (p+1)n^p &< (n+1)^{p+1} - n^{p+1}\end{align*}$$

which completes the proof.

To prove: $(n+1)^{p+1}-n^{p+1} < (n+1)^p$

Proof (direct):

By the definition of $a$ and $b$ previously given and from inequality $(1)$ we have

$$\frac{a^p}{a^k} < \frac{b^p}{b^k}$$

Let multiplying both sides by $b^k$

$$a^{p-k}b^k < b^p$$

Let take the sum of both sides

$$\sum_{k=0}^pa^{p-k}b^k < \sum_{k=0}^pb^p$$

On the RHS $b^p$ is summed $p+1$ times. Then

$$\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p$$

Let multiply the LHS by $1 = b-a$

$$\begin{align*}(b-a)\sum_{k=0}^pa^{p-k}b^k < (p+1)b^p\\ \sum_{k=0}^p[a^{p-k}b^{k+1}-a^{p-(k-1)}b^k] < (p+1)b^p\end{align*}$$

And by applying the telescoping property for sums we get

$$\begin{align*}a^{p-p}b^{p+1}-a^{p-(0-1)}b^0 &< (p+1)b^p\\ b^{p+1}-a^{p+1} &< (p+1)b^p\\ (n+1)^{p+1}-n^{p+1} &< (p+1)(n+1)^p\end{align*}$$

which completes the proof.

It looks like I can go to sleep without remorse, doesn't it? (:


3 Answers 3


$$b^{p+1} - a^{p+1} = (b-a)(b^{p}+ b^{p-1}a+b^{p-2}a^{2}+ \cdots+a^{p})$$ Substitute $$b=n+1,\,\,\,\,\,\,\,a=n$$ Then $$(n+1)^{p+1}-n^{p+1}=(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p$$ Since $$n<n+1$$ we have $$(p+1)n^p<(n+1)^p+n(n+1)^{p-1}+n^2(n+1)^{p-2}+\cdots+n^p<(p+1)(n+1)^p.$$ Hence we are done.


As you have realised, what you need to prove is equivalent to proving $$(p+1)n^p<[(n+1)^p+(n+1)^{p-1}n+\ldots+(n+1)n^{p-1}+n^p]<(p+1)n^{p+1}$$ We'll deal with these inequalities separately. Since $n+1>n$, each term inside the brackets is greater than to $n^p$ (with the exception of the final term, which is equal to $n^p$). There are $p+1$ terms inside the brackets, so the sum is greater than $(p+1)n^p$. Now we do the same argument, except the other way round, i.e. $n<n+1$ so each term in the sum is less than $(n+1)^p$ except the first. Hence the sum is less than $(p+1)(n+1)^p$.

  • $\begingroup$ That's quite intuitive. Although I don't get that of the upper bound of $(p+1)(n+1)^p$ being equivalent to $(p+1)n^{p+1}$ |:. $\endgroup$
    – asd
    Feb 11, 2015 at 13:41
  • $\begingroup$ I don't really understand your question, but your new proof is correct, if a little long-winded. $\endgroup$
    – Jason
    Feb 12, 2015 at 1:11

The inequality $\dfrac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p+1}$ is trivial. You obtain the left-hand-side by subtracting from the right and dividing by a number larger than $1$.

On the other hand, if you use part (a) you have $$(n+1)^{p+1} - n^{p+1} = (1) \bigg((n+1)^p + (n+1)^{p-1}n + \cdots + (n+1)n^{p-1} + n^p\bigg).$$ Each term on the right is $\ge n^p$, and there are $p+1$ of them. Thus $$(n+1)^{p+1} - n^{p+1} >(p+1)n^p.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .