Series Question IIT JAM 2018

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General Mathematics Question: 


Solution: 

$\sum_{n=1}^\infty (-1)^{n+1}\frac{a_{n+1}}{n!}$

$=\sum_{n=1}^\infty (-1)^{n+1}\frac{n+1+\frac{1}{n+1}}{n!}$

$=\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!(n+1)}$

$=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n-1)!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n+1)!}$

$=[1-\frac{1}{1!}+\frac{1}{2!}-...]+ [1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$

$=e^{-1}+[1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$

$=e^{-1}+[\mathbf{1-1}+1-\frac{1}{2!}+\frac{1}{3!}-...]+[\mathbf{1-\frac{1}{1!}}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$

$=e^{-1}+[1-(1-\frac{1}{1!}+\frac{1}{2!}-...)]+e^{-1}$

$=e^{-1}+[1-e^{-1}]+ e^{-1}$

$=e^{-1}+1$

Hence answer is option (D)

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