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  4. If Sn=(1+3-1)(1+3-2)(1+3-4)(1+3-8) (1+3-2n), then S ∞ is equal to

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Q. If $S_{n}=\left(1+3^{-1}\right)\left(1+3^{-2}\right)\left(1+3^{-4}\right)\left(1+3^{-8}\right) \quad\left(1+3^{-2^{n}}\right)$, then $S _{\infty}$ is equal to

Sequences and Series

Solution:

$S_{n}=\left(1+3^{-1}\right)\left(1+3^{-2}\right)\left(1+3^{-4}\right)\left(1+3^{-4}\right) \ldots\left(1+3^{-2^{n}}\right)$
$\Rightarrow \left(1-3^{-1}\right) S_{n}$
$=\left(1 3^{-1}\right)\left(1+3^{-1}\right)\left(1+3^{-2}\right)\left(1+3^{-4}\right)\left(1+3^{-8}\right) \ldots\left(1+3^{-2^{n}}\right)$
$\Rightarrow \frac{2}{3} S_{n}=\left(1-3^{-2}\right)\left(1+3^{-2}\right)\left(1+3^{-4}\right)\left(1+3^{-8}\right) \ldots .\left(1+3^{-2 n}\right)$
$=\left(1-3^{-4}\right)\left(1+3^{-4}\right)\left(1+3^{-8}\right) \ldots\left(1+3^{-2^{n}}\right)$
$=\left(1-3^{-8}\right)\left(1+3^{-8}\right) \ldots\left(1+3^{-2^{n}}\right)$
......................................................
$=\left(1-3^{-2^{n}}\right)\left(1+3^{-2^{n}}\right)=1-\left(3^{-2^{n}}\right)^{2}=1-3^{-2^{n+1}}$
$\Rightarrow S_{n}=\frac{3}{2}\left(1-3^{-2^{n+1}}\right) $
$ \therefore S_{\infty}=\frac{3}{2}(1-0)=\frac{3}{2}$