Sum it up!

Nice question! The key here is to rewrite $S$ . Let $f(n) = \displaystyle \sum_{d|n!} \dfrac{1}{d+\sqrt{n!}} \implies 2f(n) = 2\displaystyle \sum_{d|n!} \dfrac{1}{d+\sqrt{n!}}.$ Now since $\displaystyle \sum_{d|n!} \dfrac{1}{d+\sqrt{n!}} = \displaystyle \sum_{d|n!} \dfrac{1}{\dfrac{n!}{d}+\sqrt{n!}} ,$ it is easy to see that $2f(n) = \displaystyle \sum_{d|n!} \left (\dfrac{1}{d+\sqrt{n!}} + \dfrac{1}{\dfrac{n!}{d}+\sqrt{n!}} \right) .$ Then by simplification $2f(n) =\displaystyle \sum_{d|n!} \left( \dfrac{1}{d+\sqrt{n!}} + \dfrac{1}{\dfrac{n!}{d}+\sqrt{n!}}\right) =\sum_{d|n!} \dfrac{(d+\dfrac{n!}{d}) + 2\sqrt{n!}}{d\dfrac{n!}{d} +(d+\dfrac{n!}{d})\sqrt{n!} + n!} =\sum_{d|n!} \dfrac{(d+\dfrac{n!}{d}) + 2\sqrt{n!}}{2\cdot n! +(d+\dfrac{n!}{d})\sqrt{n!} } =\sum_{d|n!} \dfrac{1}{\sqrt{n!}}.$ Thus, $f(n) = \displaystyle \sum_{d|n!}\dfrac{1}{2\cdot\sqrt{n!}} =\dfrac{1}{2\cdot\sqrt{n!}} \displaystyle \sum_{d|n!}1 .$ Finally, $f(n) = \dfrac{\sigma (n!)}{2\sqrt{n!}}$ and $f(10) = \dfrac{\sigma (10!)}{2\sqrt{10!}} = \dfrac{(8+1)(4+1)(2+1)(1+1)}{2\sqrt{10!}} = \dfrac{135}{\sqrt{10!}}.$