I need to find the sum for any $p \in \mathbb{R}$, $$S=\lim\limits_{n \to \infty} \sum\limits_{r=1}^{n}\dfrac{1}{(n+r)^p}$$
My attempt: I will use the idea that $\lim\limits_{n \to ∞} f \cdot g = \lim\limits_{n \to ∞} f \cdot \lim\limits_{n \to ∞} g$ if and only if $f\cdot g$ is not an indeterminate form as $n \to ∞$,
$$S= \lim\limits_{n \to \infty} \sum\limits_{r=1}^{n}\dfrac{1}{n^p(1+r/n)^p}= \lim\limits_{n \to ∞} \dfrac{1}{n^{p-1}} \cdot \dfrac{1}{n} \sum\limits_{r=1}^{n} \dfrac{1}{\left(1+r/n\right)^p}=\lim\limits_{n \to ∞} \dfrac{1}{n^{p-1}} \cdot \lim\limits_{n \to ∞} \dfrac{1}{n} \sum\limits_{r=1}^{n} \dfrac{1}{\left(1+r/n\right)^p}= \lim\limits_{n \to ∞} \dfrac{1}{n^{p-1}} \int_{1}^{2} \dfrac{1}{x^p} \, \mathrm{d}x$$
If $p≠1$ then, $$S= \lim\limits_{n \to ∞} \dfrac{1}{n^{p-1}} \left[\left(\dfrac{2^{-p+1}}{-p+1}\right)-\left(\dfrac{1^{-p+1}}{-p+1}\right) \right]$$
If $p=1$ then, $$S= \lim \limits_{n\to ∞} \dfrac{1}{n^{1-1}} \cdot \ln 2$$
Conclusion: If $p>1$ then $S=0$, if $p=1$ then $S= \ln 2$ and if $p<1$ then $S = +\infty$
Is the reasoning correct at each and every step? Secondly, are there any alternative methods to do this? I couldn't find this question on the site.
Best Answer
The result is correct. The way of writing it, not really, when $p\neq1$. I would compute the second limit, i.e. $$\lim_{n\to+\infty}\frac{1}{n}\sum_1^n\frac{1}{(1+r/n)^p}=\int_0^1\frac{dx}{(1+x)^p}=c_p,$$ with $c_p>0$ the constant that you have obtained, and then conclude that $\frac{1}{n^{p-1}}\frac{1}{n}\sum_1^n\frac{1}{(1+r/n)^p}$ behaves as $c_pn^{1-p}$, hence the result.
Another way of arguing (when $p\neq1$) is by bunding the terms of the sum, form bellow or from above, depending on whether $p<1$ or not.
If $p>1$, then $$0 \leq \sum_{r=1}^n\frac{1}{(n+r)^p} \leq \sum_{r=1}^n\frac{1}{n^p} = \frac{n}{n^p} = n^{1-p},$$ hence the limit is $0$.
If $p<1$, then $$\sum_{r=1}^n\frac{1}{(n+r)^p} \geq \sum_{r=1}^n\frac{1}{(n+n)^p} = \frac{n^{1-p}}{2^p},$$ so the limit is $+\infty$.