Asymptotics (big-O) for power sum

Question

I am trying to prove the following equality, for $\alpha <-1$ and $x \geq 1$ $$\sum_{n \leq x}n^\alpha=\sum_{n=1}^\infty n^\alpha+\mathcal{O}(x^{\alpha+1})$$ and have tried rearranging $$\sum_{n \leq x}n^\alpha-\sum_{n=1}^\infty n^\alpha=\mathcal{O}(x^{\alpha+1})\implies \left|\sum_{n \geq x+1}n^\alpha\right| \leq cx^{\alpha+1}, \ c >0$$ I know that the series in the RHS converges by the $p$-series test. I am not sure I can use the Euler-Maclaurin formula since I would be required to use improper integrals to bound the sum. Am I missing something here? Thanks!

Answer 1

You like to prove that for $x\geq 1$ and $\alpha<-1$ $$ \sum\limits_{n > x} {n^\alpha } = \mathcal{O}(x^{\alpha + 1} ). $$ We can bound the sum via an integral as follows. If $\left[ x \right]$ denotes the integer part of $x$ then $$ \sum\limits_{n > x} {n^\alpha } \le \left[ x \right]^\alpha + \sum\limits_{n \ge \left[ x \right] + 1} {n^\alpha } \le \left[ x \right]^\alpha + \sum\limits_{n \ge \left[ x \right] + 1} {\int_{n - 1}^n {t^\alpha dt} } = \left[ x \right]^\alpha + \int_{\left[ x \right]}^{ + \infty } {t^\alpha dt} \\ = \left[ x \right]^\alpha - \frac{{ \left[ x \right]^{\alpha + 1} }}{{\alpha + 1}} = \mathcal{O}(\left[ x \right]^\alpha ) + \mathcal{O}(\left[ x \right]^{\alpha + 1} ) = \mathcal{O}(x^\alpha ) + \mathcal{O}(x^{\alpha + 1} ) = \mathcal{O}(x^{\alpha + 1} ). $$