Sum of Reciprocal of a sum of square roots

Question

I am trying to find a closed form expression of the following sum:

$\sum_{k=1}^{N}\frac{1}{\sqrt{k}+\sqrt{k+3}},\; N>1.$

I tried to determine whether methods used for evaluating more conventional sums (e.g. sum of integers, sum of powers of integers, etc) could be adapted here, but with no success. Moreover, I am looking for a method that can be applied to a generalization of this problem I thought of:

$\sum_{k=1}^{N}\frac{1}{\sqrt{k+m}+\sqrt{k+n}}, \; n, m, \in \mathbb{Z}; \; n>m>0, \; N>1.$

For this second sum, I am likewise interested in an equivalent expression; I am not sure if the expression may have to be written in sigma/product notation, although in the spirit of the first problem, the index should only depend on the parameters $m$ and $n$, not $N$.

Answer 1

Note that $$\dfrac1{\sqrt{k}+\sqrt{k+3}} = \dfrac{\sqrt{k+3}-\sqrt{k}}3$$ Hence, $$\sum_{k=1}^N \dfrac1{\sqrt{k}+\sqrt{k+3}} = \sum_{k=1}^N \dfrac{\sqrt{k+3}-\sqrt{k}}3 = \dfrac13 \left(\sqrt{N+3}+\sqrt{N+2}+\sqrt{N+1}-\sqrt3 -\sqrt2-1\right)$$ Similarly, we can obtain an expression for $\displaystyle \sum_{k=1}^N \dfrac1{\sqrt{k+m}+\sqrt{k+n}}$

Answer 2

A generalisation can be put forward as follows: $$\sum_{k=1}^{N}\frac{1}{\sqrt{k+m}+\sqrt{k+n}}$$ $$=\sum_{k=1}^{N}\frac{1}{\sqrt{k+m}+\sqrt{k+n}}\cdot \frac{\sqrt{k+m}-\sqrt{k+n}}{\sqrt{k+m}-\sqrt{k+n}}$$ $$=\sum_{k=1}^{N}\frac{\sqrt{k+m}-\sqrt{k+n}}{k+m-k-n}$$ $$=\frac{1}{m-n} \sum_{k=1}^{N}(\sqrt{k+m}-\sqrt{k+n})$$ $$=\frac{1}{n-m} \sum_{k=1}^{N}(\sqrt{k+n}-\sqrt{k+m})$$ $$=\frac{1}{n-m} \left(\sqrt{N+n}+\sqrt{N+n-1}+\ldots +\sqrt{N+m+1}-\sqrt{n}-\sqrt{n-1}-\dots -\sqrt{2+m}-\sqrt{1+m}\right)$$

Check if you can find a closed form.