How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# 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) $1+e^{-1}$

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.

# 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.