If ∑\sum∑r=1nTr_{r=1}^n T_rr=1nTr = (2n−1)(2n+1)(2n+3)(2n+5)64\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}64(2n−1)(2n+1)(2n+3)(2n+5) , then limn→∞∑r=1n1Tr \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} limn→∞∑r=1nTr1 is equal to :