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If $L = lim_{n \to \infty} (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) (1 + 3^{- 8}) \dots$ $(1 + 3^{- 2^{n}}),$ then

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Correct option is B

1−3−1L=limn→∞ 1−3−11+3−11+3−21+3−4…1+3−2n 23L=limn→∞ 1−3−21+3−21+3−4…1+3−2n=limn→∞ 1−3−221+3−4…1+3−2n=limn→∞ 1−3−2n+1=1⇒L=3/2

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If L=limn→∞ 1+3−11+3−21+3−41+3−8… 1+3−2n, then

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If $L = lim_{n \to \infty} (1 + 3^{- 1}) (1 + 3^{- 2}) (1 + 3^{- 4}) (1 + 3^{- 8}) \dots$ $(1 + 3^{- 2^{n}}),$ then