Tight upper bound on $\sum_{1\leq k\leq m} \sqrt{(n/k)+b}$

Question

I apologise if this is obvious or off-topic.

Let $n$ be large and fixed, $b>0,$ and $m \in [\log n,n),$ say. This sum seems to be hard to evaluate/upper bound analytically (in closed form). Approximating by integration is tricky. To use a $u-$substitution for teh quantity inside the square root, would mean that the estimated integral would have an extra logarithm which would mean the bound wouldn't be so great. Mathematica (via Wolfram Alpha) gives (after factoring out $\sqrt{n}$):

but the problem here is that the inverse hyperbolic tangent blows up if its argument is strictly bigger than 1.

It doesn't help that all my books, including Gradsteyn-Rhzik are at work, and so is my licenced copy of Mathematica--off limits due to COVID-19.

Any comments, suggestions appreciated. Maybe there is something obvious I haven't seen.

Answer 1

Assuming the $x$ in your title is the same as $k$, you want to upper-bound $$s:=\sum_{k=1}^m\sqrt{n/k+b}=\sqrt n\,\sum_{k=1}^m f(k),$$ where $$f(x):=\sqrt{1/x+c},\quad c:=b/n>0.$$ The function $f$ is strictly convex on $(0,\infty)$, whence $$s=\sqrt n\,\sum_{k=1}^m f(k) \\ <\sqrt n\,\int_{1/2}^{m+1/2}f(x)\,dx \\ =\sqrt n\,\Big( \frac{\sqrt{(2 m+1) (2 c m+c+2)}}2-\frac{\sqrt{c+2}}{2}\Big) \\ +\frac{\sqrt n}{2\sqrt{c}}\, \ln\frac{(2 m+1) \left(\sqrt{\frac{2}{2 c m+c}+1}+1\right)^2}{\left(\sqrt{\frac{c+2}{c}}+1\right)^2}.$$