∑n=0∞(logex)2n(2n)!=

∑n=0∞ (log x)2n2n =1+(log x)22+(log x)44+.... =elog x+e-log x2=x+1x2=x2+12x

∑n=0∞(logex)2n(2n)!=

∑n=0∞ (log x)2n2n =1+(log x)22+(log x)44+.... =elog x+e-log x2=x+1x2=x2+12x

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$\sum_{n = 0}^{\infty} \frac{{(\log_{e} x)}^{2 n}}{(2 n)!} =$

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x + 1

x²+1

$\frac{x^{2} + 1}{2 x}$

x – 1

answer is D.

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Detailed Solution

$\sum_{n = 0}^{\infty} \frac{{(\log x)}^{2 n}}{2 n} = 1 + \frac{{(\log x)}^{2}}{2} + \frac{{(\log x)}^{4}}{4} + . . . . = \frac{e^{\log x} + e^{- \log x}}{2} = \frac{x + \frac{1}{x}}{2} = \frac{x^{2} + 1}{2 x}$

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