Understand the problem
Let $a_{n}=n+\frac{1}{n}, n \in \mathbb{N} .$ Then the sum of the series $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{a_{n+1}}{n !}$ is -(A) $e^{-1}-1$
(B) $e^{-1}$
(C) $1-e^{-1}$
(D) $1+e^{-1}$Start with hints
Do you really need a hint? Try it first![Hint 1] Consider $a_n = n + \frac{1}{n} , n \in \mathbb{N} $
We have to use $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!}+.....$
Specifically $e^{1} = 1+ \frac{1}{1!} + \frac{1}{2!}+....$
And $ e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + .......$
Do you want to play with it.[Hint 2]$ \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{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]$
Now we will be breaking it term by term for the ease of calculation. Can you do it from here?[Hint 3]$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} $
= $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$
= $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1] $ = $e^{-1} + 1 $ So option (D) is our required answer.Let's have a look into the graphs
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