Upper bound for the sum of $~i^{-1.5}~$

Question

For some $n>0$, can we obtain an upper bound of $$f(n):=\sum_{i=n}^{\infty}\frac{1}{i^{1.5}}~?$$

In my understanding, a constant $C$ exist such that $f(n)\leq C$.

Are there better bounds like $f(n)=O(n^{-0.5})$?

Answer 1

Well, $\sum_{n\geq 1}\frac{1}{n\sqrt{n}}$ is a constant, $\zeta\left(\frac{3}{2}\right)=2.6123753\ldots$, and $f(x)=\frac{1}{x\sqrt{x}}$ is a convex function on $\mathbb{R}^+$.

By the Hermite-Hadamard inequality it follows that $$ \sum_{n\geq N}\frac{1}{n\sqrt{n}}\leq\int_{N-1/2}^{+\infty}\frac{dx}{x\sqrt{x}}= \frac{2}{\sqrt{N-\frac{1}{2}}}=O\left(\frac{1}{\sqrt{N}}\right).$$

Addendum: if you need a tighter bound, you may prove by creative telescoping that for any $N\geq 1$ $$ \sum_{n\geq N}\frac{1}{n\sqrt{n}}\leq \frac{2}{\sqrt{N-\frac{1}{2}+\frac{1}{16N}}} $$ holds. $\zeta\left(\frac{3}{2}\right)<1+\frac{8}{7}\sqrt{2}$ is an interesting consequence.