# Proving two inequalities involving integrals and sums

Let $$p \ge 1$$. Show that $$$$\frac{1}{(k+1)^p} \le \int_{k}^{k+1} \frac{1}{x^p} \,dx \le \frac{1}{k^p}$$$$ for every positive integer $$k$$. Hence show that $$$$\sum_{k=2}^{n+1} \frac{1}{k^p} \le \int_{1}^{n+1} \frac{1}{x^p} \,dx \le \sum_{k=1}^{n} \frac{1}{k^p}$$$$ for every positive integer $$n$$.

My approach to the first part is simple: evaluate the integral $$\int_{k}^{k+1} \frac{1}{x^p} \,dx$$, in which I generated $$-\frac{1}{(p-1)(k+1)^{p-1}}+\frac{1}{(p-1)k^{p-1}}$$. I tried to expand this by multiplying the denominator on the three components of the inequality. However, it seems tedious and I could not prove the inequalities by such methods.

Are there any better alternatives proving the two inequalities? Hints/suggestions/complete answers are all appreciated. Thanks.

• Hint: for $x\in[k,k+1]$, $\displaystyle{\frac{1}{(k+1)^p}\le\frac{1}{x^p}\le\frac{1}{k^p}}$. Apr 15, 2022 at 7:01
• @Jean-ClaudeArbaut nice hint, I can grasp your logic behind :) Apr 15, 2022 at 7:15

To begin with, we note that for every positive integer $$k$$, $${1 \over (k + 1)^p} \leq {1 \over x^p} \ \leq {1 \over k^p}, \ \ \mbox{where} \ \ k \leq x \leq k + 1 \tag{1}$$
Integrating (1) from $$k$$ to $$k + 1$$, it is immediate that $${1 \over (k + 1)^p} \leq \int\limits_{k}^{k + 1} \ {1 \over x^p} \ dx\leq {1 \over k^p}, \ \ \mbox{where} \ \ k \leq x \leq k + 1 \tag{2}$$
For each integer $$k$$ from $$1$$ to $$n$$, we can write the following inequalities: $${1 \over 2^p} \leq \int\limits_{1}^{2} \ {1 \over x^p} \ dx \leq {1 \over 1^p} \tag{s1}$$ $$\vdots$$ $${1 \over (n+1)^p} \leq \int\limits_{n}^{n+1} \ {1 \over x^p} \ dx \leq {1 \over n^p} \tag{sn}$$
Adding the inequalities (s1) to (sn), the result follows, viz. $$\sum\limits_{k = 2}^{n + 1} \ {1 \over k^p} \leq \int\limits_{1}^{n+1} \ {1 \over x^p} \ dx \leq \sum\limits_{k = 1}^{n} \ {1 \over k^p}$$
Fix $$p\ge 1$$. Consider $$f(x) = x^{-p}$$ defined for all $$x > 0$$. Then, $$f'(x) = -px^{-p-1} < 0$$ for all $$x > 0$$, i.e., the function $$f$$ is monotonically decreasing for all positive $$x$$. In particular, choose $$k\in \mathbb N$$ and consider the interval $$[k,k+1]$$. For any $$x \in [k,k+1]$$, we have $$f(k+1) \le f(x) \le f(k)$$ i.e., $$\frac{1}{(k+1)^p} \le \frac{1}{x^p} \le \frac{1}{k^p}$$ Integrating over $$[k,k+1]$$, we have $$\int_k^{k+1}\frac{1}{(k+1)^p}\, \mathrm{d}x \le \int_k^{k+1}\frac{1}{x^p}\, \mathrm{d}x \le \int_k^{k+1}\frac{1}{k^p}\, \mathrm{d}x$$ which is just $$\frac{1}{(k+1)^p} \le \int_k^{k+1}\frac{1}{x^p}\, \mathrm{d}x\le \frac{1}{k^p}$$ This establishes the first inequality for all $$p\ge 1$$ and all $$k\in \Bbb N$$. I hope you can take it from here.