Proving two inequalities involving integrals and sums 
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.
 A: 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.
$$
\sum\limits_{k = 2}^{n + 1} \ {1 \over k^p} \leq
\int\limits_{1}^{n+1} \ {1 \over x^p} \ dx \leq \sum\limits_{k = 1}^{n} \ {1 \over k^p}
$$
A: 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.
