Q.If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64},$ then $\lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_{r}} \right)$ is equal to:

\frac{2}{3}

\frac{1}{3}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.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},∑r=1n​Tr​=64(2n−1)(2n+1)(2n+3)(2n+5)​, thenlim⁡n→∞∑r=1n(1Tr) \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_{r}} \right)limn→∞​∑r=1n​(Tr​1​) is equal to: