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MCQMediumJEE Main 2025 · 22 January, Shift 1Sum of Series

Mathematics Question from JEE Main 2025 · 22 January, Shift 1

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

  • A

    11

  • B

    00

  • C

    23\frac{2}{3}

  • D

    13\frac{1}{3}

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