The value of \underset{n \rightarrow \infty}{Lim} \sum_{r = 1}^{n} \frac{6 n - 1}{\left(\left(\right. 3 n - 2 \left.\right)\right)^{2} \cdot \left(\left(\right. 3 n + 1 \left.\right)\right)^{2}} is L, then 9L is equal to

Engineering

Mathematics

Special Types of Sequence

Question

The value of $\underset{n \to \infty}{Lim} \sum_{r = 1}^{n} \frac{6 n - 1}{{(3 n - 2)}^{2} \cdot {(3 n + 1)}^{2}}$ is L, then 9L is equal to

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$L = \frac{1}{3} \underset{n \to \infty}{Lim} \sum_{r = 1}^{n} \frac{{(3 n + 1)}^{2} - {(3 n - 2)}^{2}}{{(3 n - 2)}^{2} \cdot {(3 n + 1)}^{2}} = \frac{1}{3} \underset{n \to \infty}{Lim} \sum_{r = 1}^{n} (\frac{1}{{(3 n - 2)}^{2}} - \frac{1}{{(3 n + 1)}^{2}}) = \frac{1}{3}$