Greatest Integer Function - A mess!

Use Euclid algorithm couple of times $d_N=((N+1)^2+20,N^2+20) \implies d_N=(N^2+20,2N+1)$

Note that $d_N| 2N+1$ , since $d_N$ is the GCD of $2N+1$ and another number. Therefore the fractional part is $0$ .

Since $d_n = \gcd(a_n, a_{n+1})$ , we can write $a_n = d_n b_n$ and $a_{n+1} = d_n b_{n+1}$ . where $b_n$ and $b_{n+1}$ are positive integers, and $b_{n+1} > b_n$ . Then we have:

$\frac {2n+1}{d_n} = \frac {(n+1)^2+20 - (n^2+20)}{d_n} = \frac {a_{n+1}-a_n}{d_n} = \frac {d_n(b_{n+1}-b_n)}{d_n} = b_{n+1} - b_n$

Therefore $\left \{ \dfrac {2n+1}{d_n} \right \} = \{b_{n+1} - b_n \} = \boxed 0$ .