How can we show that for every $p > 1$, we have $\sum_{n>x} n^{-p} \ll_p x^{1 - p}$ for $x\geq 1$

Question

How can we show that for every $p > 1$, we have $\sum_{n>x} n^{-p} \ll_p x^{1 - p}$ for $x\geq 1$? I notice that LHS $\leq \sum_{k \geq 0}(x + k)^{-p}$, but I don't know what to do next.

Answer 1

For $x\geq1$ and $p>1$, we have \begin{align*} \sum_{n=\lceil x\rceil}^\infty\frac1{n^p}&=\sum_{n=\lceil x\rceil}^\infty\int_n^\infty\!\!\frac p{t^{p+1}}dt\\[.5em] &=\int_{\lceil x\rceil}^\infty\sum_{n=\lceil x\rceil}^{\lfloor t\rfloor}\frac p{t^{p+1}}dt\\[.5em] &=p\int_{\lceil x\rceil}^\infty\frac{\lfloor t\rfloor-\lceil x\rceil+1}{t^{p+1}}dt\\[.5em] &\leq p\int_x^\infty\frac1{t^p}dt\\[.5em] &=\frac p{p-1}x^{1-p}. \end{align*}

The second equality is by interchanging the sum and integral (similar as changing the order of a double integral).