Proving two inequalities involving integrals and sums

Question

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

Answer 1

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. $$ \sum\limits_{k = 2}^{n + 1} \ {1 \over k^p} \leq \int\limits_{1}^{n+1} \ {1 \over x^p} \ dx \leq \sum\limits_{k = 1}^{n} \ {1 \over k^p} $$

Answer 2

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.