Question:
If $0
Correct Option: 1
Solution:
$y=\left(1-\frac{1}{2}\right) x^{2}+\left(1-\frac{1}{3}\right) x^{3}+\ldots .$
$=\left(x^{2}+x^{3}+x^{4}+\ldots \ldots\right)-\left(\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+\ldots .\right)$
$=\frac{x^{2}}{1-x}+x-\left(x+\frac{x^{2}}{2}+\frac{x^{2}}{3}+\ldots\right)$
$=\frac{x}{1-x}+\ell n(1-x)$
$x=\frac{1}{2} \Rightarrow y=1-\ell \mathrm{n} 2$
$e^{1+y}=e^{1+1-\ell n 2}$
$=\mathrm{e}^{2-\ell \mathrm{n} 2}=\frac{\mathrm{e}^{2}}{2}$
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