lim _{n rightarrow infty} left( frac{1}{3 cdo | vedclass

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(A) The general term of the series is $T_k = \frac{1}{(4k-1)(4k+3)}$.We can write $T_k = \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)$.T

Vedclass · Accepted Answer

$\frac{1}{12}$

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(A) The general term of the series is $T_k = \frac{1}{(4k-1)(4k+3)}$.We can write $T_k = \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)$.This is a telescoping series:$S_n = \frac{1}{4} \left[ \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{11} \right) + \ldots + \left( \frac{1}{4n-1} - \frac{1}{4n+3} \right) \right]$.$S_n = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{4n+3} \right)$.Taking the limit as $n \rightarrow \infty$:$\lim _{n \rightarrow \infty} S_n = \lim _{n \rightarrow \infty} \frac{1}{4} \left( \frac{1}{3} - \frac{1}{4n+3} \right) = \frac{1}{4} \left( \frac{1}{3} - 0 \right) = \frac{1}{12}$.

$\lim _{n \rightarrow \infty} \left( \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \ldots + n \text{ terms} \right) =$

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