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Question: \[\frac{1}{2} + \frac{1}{4} + \frac{1}{8(2)!} + \frac{1}{16(3)!} + \frac{1}{32(4)!} + ......\infty =...

12+14+18(2)!+116(3)!+132(4)!+......∞=\frac{1}{2} + \frac{1}{4} + \frac{1}{8(2)!} + \frac{1}{16(3)!} + \frac{1}{32(4)!} + ......\infty =

A

ee

B

e\sqrt{e}

C

e2\frac{\sqrt{e}}{2}

D

None of these

Answer

e2\frac{\sqrt{e}}{2}

Explanation

Solution

12+log⁡e(23)\frac{1}{2} + \log_{e}\left( \frac{2}{3} \right).