Are you as smart as Gauss?

We first calculate $d(11!)$ , the number of positive divisors of $11!$ , by finding its prime decomposition. $11!$ will only have prime factors $2, 3, 5, 7$ and $11$ ; if we write $11! = 2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e$ , we can find the exponents of each prime factor as follows:

$\begin{array}{rll} a &= \ \ \left\lfloor \dfrac{11!}{2} \right\rfloor + \left\lfloor \dfrac{11!}{2^2} \right\rfloor + \left\lfloor \dfrac{11!}{2^3} \right\rfloor \ \ = \ \ 8 \\ \\ b &= \ \ \left\lfloor \dfrac{11!}{3} \right\rfloor + \left\lfloor \dfrac{11!}{3^2} \right\rfloor \ \ = \ \ 4 \\ \\ c &= \ \ \left\lfloor \dfrac{11!}{5} \right\rfloor \ \ = \ \ 2 \\ \\ d &= \ \ \left\lfloor \dfrac{11!}{7} \right\rfloor \ \ = \ \ 1 \\ \\ e &= \ \ \left\lfloor \dfrac{11!}{11} \right\rfloor \ \ = \ \ 1 \end{array}$

This gives $11! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11$ , and then the number of positive divisors would be $d(11!) = (8 + 1)(4 + 1)(2 + 1)(1 + 1)(1 + 1) = 540$ .

We now arrange the factors, not in increasing size, but by factor pairs; i.e. $d_1 = 1, d_2 = 11!, d_3 = 2, d_4 = \frac{11!}{2}, \dots$ Then $d_{2k - 1}$ and $d_{2k}$ will form factor pairs, and we will have 270 such factor pairs.

Consider any two terms in the sum formed from such a factor pair.

$\begin{array}{rlccl} \dfrac{1}{d_{2k - 1}+\sqrt{11!}} + \dfrac{1}{d_{2k} + \sqrt{11!}} &= \ \ \dfrac{d_{2k} + \sqrt{11!} + d_{2k - 1}+\sqrt{11!}}{\left( d_{2k - 1} + \sqrt{11!} \right) \left(d_{2k} + \sqrt{11!} \right)} \\ \\ &= \ \ \dfrac{2 \sqrt{11!} + d_{2k - 1} + d_{2k}}{d_{2k - 1} d_{2k} + \sqrt{11!} \left( d_{2k - 1} + d_{2k} \right) + 11!} & & & \small \text{[ But } d_{2k - 1} d_{2k} = 11! \ ] \\ \\ &= \ \ \dfrac{2 \sqrt{11!} + d_{2k - 1} + d_{2k}}{2 (11!) + \sqrt{11!} \left( d_{2k - 1} + d_{2k} \right)} \\ \\ &= \ \ \dfrac{2 \sqrt{11!} + d_{2k - 1} + d_{2k}}{\sqrt{11!} \left( 2 \sqrt{11!} + d_{2k - 1} + d_{2k} \right)} \\ \\ &= \ \ \dfrac{1}{\sqrt{11!}} \end{array}$

Thus every such pair of terms adds up to $\dfrac{1}{\sqrt{11!}}$ , and there are $270$ such pairs in our sum. Therefore

$\begin{array}{rl} \sum_{k = 0}^n\ \frac{1}{d_k + \sqrt{11!}} &= \ \ \dfrac{270}{\sqrt{11!}} \\ \\ &\approx \ \ \boxed{0.042735216} \end{array}$