Doubt from series

Question 69

-1,i.e option B ?

Yes correct i think i made a mistake i had got zero

\begin{align*} p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ & = \frac{1}{2}\left(1 + \frac{1}{k}\right) \end{align*}

As {k\to\infty}

\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{k\to\infty} p_{k} = \frac{1}{2}.

Therefore,\log_{2}{\frac{1}{2}}=-1

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Are these questions from the NTA Tests @abhijit_2020 ?

No mypat tests.