In the proof of PNT by Don Zagier, he wrote below lines-
The series on the right (which is $\sum_{n=1}^\infty \int_{n}^{n+1} \left(\frac{1}{n^s}-\frac{1}{x^s}\right) dx$) converges absolutely for $R(s)>0$ because-
$$ \left |\int_{n}^{n+1} \left(\frac{1}{n^s}-\frac{1}{x^s}\right) dx\right|= \left |s\int_{n}^{n+1} \int_{n}^{x} \frac{du}{u^{s+1}} dx\right| \le \frac{|s|}{n^{R(s)+1}}$$ by the mean value theorem.
I don't understand how the mean value theorem implies the convergence, can anyone please explain? I am new to the topic. Thanks!
The source of the question can be found here.
$\left |\int_{n}^{n+1} \left(\frac{1}{n^s}-\frac{1}{x^s}\right) dx\right|= \left |s\int_{n}^{n+1} \int_{n}^{x} \frac{du}{u^{s+1}} dx\right| \le \frac{|s|}{n^{R(s)+1}} $
Since $u^{s+1} \ge n^{s+1} $ for $u \ge n$,
$\begin{array}\\ \int_{n}^{x} \frac{du}{u^{s+1}} &\le \int_{n}^{x} \frac{du}{n^{s+1}} \\ &= \frac{x-n}{n^{s+1}} \\ \text{so}\\ \int_{n}^{n+1} \int_{n}^{x} \frac{du}{u^{s+1}} dx &\le \int_{n}^{n+1} \frac{x-n}{n^{s+1}} dx\\ &=\dfrac1{n^{s+1}} \int_{0}^{1} x dx\\ &=\dfrac1{2n^{s+1}} \\ \end{array} $
You can put in absolute value signs for complex $s$.