Find $\lfloor S \rfloor$

Note that $a_2=\tfrac34$ , so that $S > \frac{1}{a_1+1} + \frac{1}{a_2+1} = \frac{26}{21} > 1$ so that $\lfloor S \rfloor \ge 1$ . More generally, $\frac{1}{a_n} - \frac{1}{a_{n+1}} \; = \; \frac{a_{n+1} - a_n}{a_{n+1}a_n} \; = \; \frac{a_n^2}{a_{n+1}a_n} \; = \; \frac{a_n}{a_{n+1}} \; = \; \frac{1}{a_n + 1 }$ and hence $\sum_{n=1}^N \frac{1}{a_n+1} \; = \; \sum_{n=1}^N\left(\frac{1}{a_n} - \frac{1}{a_{n+1}}\right) \; = \; \frac{1}{a_1} - \frac{1}{a_{N+1}} \; = \; 2 - \frac{1}{a_{N+1}} < 2$ for all integers $N$ . Putting $N=2017$ , we have $S < 2$ , and hence $\lfloor S \rfloor = \boxed{1}$ .

Of course, since $a_3 = \tfrac{21}{16} > 1$ and $a_{n+1} > a_n^2$ for all $n \ge 1$ , the number $a_{2018}$ is very large, and so the difference between $S$ and $2$ is very small indeed. Indeed, since $a_n \ge \left(\tfrac{21}{16}\right)^{2^{n-3}} \hspace{2cm} n \ge 3$ we can calculate that $\log_{10}(a_{2018} + 1) \; > \; \log_{10}a_{2018} \; > \; 2^{2015} \log_{10}\tfrac{21}{16} \; \approx 4.44 \times 10^{605}$ and hence the difference between $S$ and $2$ is less than $10^{-4.4 \times 10^{605}}$