Lower bound or approximation of the partial sum $\sum_{k=a}^{n-1} \frac{1}{\sqrt{n-a} - \sqrt{k-a}}$

Question

I'm interested in an lower bound or approximation of the partial sum (for some $a \in \mathbb{N}$)

$$\sum_{k=a}^{n-1} \frac{1}{\sqrt{n-a} - \sqrt{k-a}}.$$

I am aware of the approximation of the sum $\sum_{k=1}^n \frac{1}{\sqrt{k}} \approx 2 \sqrt{n}$ (see e.g. this question). I'd guess that we could get a similar estimate / lower bound for the sum above, e.g. something like $\sqrt{n-a}$. However, I am at loss how to lower bound the sum appropriately.

Answer 1

The sum equals $S_{n-a}$, where for $n>0$ $$S_n=\sum_{k=0}^{n-1}\frac1{\sqrt{n}-\sqrt{k}}=\sum_{k=0}^{n-1}\left(\frac{2\sqrt{n}}{n-k}-\frac1{\sqrt{n}+\sqrt{k}}\right)=\color{blue}{(2H_n-T_n)\sqrt{n}},\\H_n:=\sum_{k=1}^n\frac1k,\qquad T_n:=\frac1n\sum_{k=0}^{n-1}\frac1{1+\sqrt{k/n}}.$$

A lower bound $H_n\geqslant\log n+\gamma$ is well-known, while an upper bound for $T_n$ might be $$T_n\leqslant\frac1{2n}+\int_0^1\frac{dx}{1+\sqrt{x}}=2(1-\log2)+\frac1{2n}.$$

Thus, $S_n$ is $O(\sqrt{n}\log n)$ but is not $O(\sqrt{n})$ as $n\to\infty$.