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Question

Question: \[\frac{1}{2!} + \frac{1 + 2}{3!} + \frac{1 + 2 + 3}{4!} + ......\infty =\]...

12!+1+23!+1+2+34!+......=\frac{1}{2!} + \frac{1 + 2}{3!} + \frac{1 + 2 + 3}{4!} + ......\infty =

A

ee

B

2e2e

C

e/2

D

None of these

Answer

e/2

Explanation

Solution

1n12n2+13n314n4+....\frac{1}{n} - \frac{1}{2n^{2}} + \frac{1}{3n^{3}} - \frac{1}{4n^{4}} + ....

Here loge(n+1n)\log_{e}\left( \frac{n + 1}{n} \right)

loge(nn+1)\log_{e}\left( \frac{n}{n + 1} \right).