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
Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
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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
Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
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