smc

It is well known that $T_n=\frac{n(n+1)}{2}$, so we can put this in and simplify to get that the general term of the product is $\frac{n(n+1)}{(n+2)(n-1)}.$ Now by writing out the terms, we see that almost everything cancels (telescopes), giving $\frac{3\times 1991}{1993}$ which is almost $3$, so closest to $\mathrm{2.9.}$ $\boxed{2.9}$