1-x2+x23-x34+……∞=

1-x2+x23-x34+…… =x-x22+x33-x44+....x =log(1+x)x

1-x2+x23-x34+……∞=

1-x2+x23-x34+…… =x-x22+x33-x44+....x =log(1+x)x

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$1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots \dots \infty =$

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$\log (1 + x)$

$- \log (1 + x)$

$- \frac{\log (1 + x)}{x}$

$\frac{\log_{e} (1 + x)}{x}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} 1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots \dots \\ = & \frac{x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . . .}{x} \\ = & \frac{\log (1 + x)}{x} \end{array}$

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