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Question: If \(S = \frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + .... + \infty,\) then \(e^{...

If S=11.2−12.3+13.4−14.5+....+∞,S = \frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + .... + \infty, then eS=e^{S} =

A

log⁡e(4e)\log_{e}\left( \frac{4}{e} \right)

B

4e\frac{4}{e}

C

log⁡e(e4)\log_{e}\left( \frac{e}{4} \right)

D

e4\frac{e}{4}

Answer

4e\frac{4}{e}

Explanation

Solution

log⁡e(1−1x)\log_{e}\left( 1 - \frac{1}{x} \right)

log⁡e(xx+1)\log_{e}\left( \frac{x}{x + 1} \right)

log⁡e(x+1)−log⁡e(x−1)=\log_{e}(x + 1) - \log_{e}(x - 1) =

2[x+x33+x55+......∞]2\left\lbrack x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ......\infty \right\rbrack

[x+x33+x55+......∞]\left\lbrack x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ......\infty \right\rbrack 2[1x+13x3+15x5+...∞]2\left\lbrack \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right\rbrack; [1x+13x3+15x5+...∞]\left\lbrack \frac{1}{x} + \frac{1}{3x^{3}} + \frac{1}{5x^{5}} + ...\infty \right\rbrack.