_ n → ∞ \ (2(1)/(2) - 2(1)/(3))(2(1)/(2) - 2(1)/(5)) . . . ....

Let,

$L = \lim_{n \to \infty}\left\{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)\left(2\dfrac{1}{2} - 2\dfrac{1}{5}\right) . . . . \left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)\right\}$

Now,

$2\dfrac{1}{2} - 2\dfrac{1}{3} < 1$

$2\dfrac{1}{2} - 2\dfrac{1}{5} < 1$

--------------------

$2\dfrac{1}{2} - 2\dfrac{1}{2n + 1} < 1 \forall n \in N$

And ${\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)}^{n} < \left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)\left(2\dfrac{1}{2} - 2\dfrac{1}{5}\right) . . . . \left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right) < {\left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)}^{n}$

$⇒ \lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)}^{n} < \lim_{n \to \infty}\left\{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)\left(2\dfrac{1}{2} - 2\dfrac{1}{5}\right) . . . . \left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)\right\} < \lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)}^{n}$

$⇒ \lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)}^{n} < L < \lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)}^{n}$

And $\lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)}^{n} = 0 \text{ and } \lim_{n \to \infty}{\left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)}^{n} = 0$

\left\{\text{as}2\dfrac{1}{2} - 2\dfrac{1}{3} < 1 \text{ and } 2\dfrac{1}{2} - 2\dfrac{1}{2n + 1} < 1\right\}$</p><p>Hence,$\lim_{n \to \infty}\left\{\left(2\dfrac{1}{2} - 2\dfrac{1}{3}\right)\left(2\dfrac{1}{2} - 2\dfrac{1}{5}\right) . . . . \left(2\dfrac{1}{2} - 2\dfrac{1}{2n + 1}\right)\right\} = 0