Let $ p >0 $ be a positive real. We consider the $p$-series
$$ \sum_{ n > N} \frac{1}{n^p} $$
where $ N \to \infty.$
An application of the integral test show that in the case where the series converges (namely for $ p > 1,$ then one has the estimate
$$ \sum_{ n > N} \frac{1}{n^p} \ll N^{1-p},$$
as $ N \to \infty.$
My question is the following.
Question 1: Under what assumptions one can obtain the tighter upper bound $$ \sum_{ n > N} \frac{1}{n^p} \ll N^{-p},$$
as $ N \to \infty?$
edit: After the comments of Sandeep Silwal , I have decided to change a bit the range of summation, hoping to get something better than $ \ll N^{1-p}.$ Here is the new question:
Question 2: Let $ \delta \in (0,1).$ Under what assumptions on p, is it true that
$$ \sum_{ n > N^\delta} \frac{1}{n^p} \ll N^{-p\delta} \quad ? $$
Finally, I think (if such an estimate is valid) that one could sum at the level $ n \geq h(N)$ where $h (N)$ is a positive slowly, increasing to infinity function, with $h(N) = o(N), \quad N \to \infty.$
An integral comparison shows that $$(N+1)^{1-p} \leq (p-1)\sum_{n \geq N}{n^{-p}} \leq N^{1-p}.$$
Thus $$\sum_{n \geq N^{\delta}}{n^{-p}} \sim \frac{N^{-\delta(p-1)}}{p-1} >> N^{-\delta p}.$$