The value of mathop {lim }limits_{n to infty | vedclass

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(A) Let the $n$-th term of the series be $T_k = \frac{1}{(2k - 1)(2k + 1)}$.Using partial fractions,we can write $T_k = \frac{1}{2} \left( \frac{1}{2k

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$1/2$

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(A) Let the $n$-th term of the series be $T_k = \frac{1}{(2k - 1)(2k + 1)}$.Using partial fractions,we can write $T_k = \frac{1}{2} \left( \frac{1}{2k - 1} - \frac{1}{2k + 1} ight)$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \frac{1}{2} \left[ \left( 1 - \frac{1}{3} ight) + \left( \frac{1}{3} - \frac{1}{5} ight) + \dots + \left( \frac{1}{2n - 1} - \frac{1}{2n + 1} ight) ight]$.This is a telescoping series,so $S_n = \frac{1}{2} \left( 1 - \frac{1}{2n + 1} ight)$.Taking the limit as $n 	o \infty$,we get $\lim_{n 	o \infty} S_n = \frac{1}{2} (1 - 0) = \frac{1}{2}$.

The value of $\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} \right)$ is equal to

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