16.12

The first two terms are $\lfloor{\frac{1}{1}}\rfloor$ and $\lfloor{\frac{3}{3}}\rfloor.$ These terms, of course, correspond to $n=0$ and $n=1.$ I will now show that $\lfloor{\frac{n^2+n+1}{n^3+n^2+1}}\rfloor=0$ for $n>1$ . $\lfloor{\frac{n^2+n+1}{n^3+n^2+1}}\rfloor=0\Leftrightarrow n^3+n^2+1>n^2+n+1\Leftrightarrow n^3-n>0 \Leftrightarrow (n-1)n(n+1)>0.$ We can see that for $n>1,$ each factor, namely, $n-1,n,n+1,$ are all positive. Therefore, their product is also positive. This shows $\lfloor{\frac{n^2+n+1}{n^3+n^2+1}}\rfloor=0$ for all $n>1.$ Therefore, only the first two terms contribute to the sum and the desired sum is equal to $\lfloor{\frac{1}{1}}\rfloor+\lfloor{\frac{3}{3}}\rfloor=1+1=2.$