$T(n) = 1 + 2 + 3 + ... + n$

(I already figured out this is always equal to $\frac{n(n+1)}{2}$, can I use this?)

$P(n) = \frac{T(2)}{T(2) - 1} . \frac{T(3)}{T(3)-1} . \frac{T(4)}{T(4)-1} . ... . \frac{T(n)}{T(n) - 1}$

How can I estimate $P(1991)$?