Sequences and Series Part 3

Concept:The sum is simplified using algebraic manipulation of factorials and then expressed in terms of the exponential series $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ at $x = -1$.Explanation:Write the general term: $\frac{k(k+1)}{k!} = \frac{k+1}{(k-1)!}$ (since $k! = k \cdot (k-1)!$).Then $k+1 = (k-1) + 2$, so term becomes $\frac{k-1}{(k-1)!} + \frac{2}{(k-1)!} = \frac{1}{(k-2)!} + \frac{2}{(k-1)!}$ for $k \geq 2$.For $k=1$, the term $\frac{1}{(-1)!}$ is taken as $0$ (factorial of negative integer is infinite).Thus the sum splits: $S = \sum_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!} + \sum_{k=1}^{\infty} \frac{2(-1)^{k+1}}{(k-1)!}$.Evaluate first sum: let $n = k-2$. Then $k+1 = n+3$, sum becomes $\sum_{n=0}^{\infty} \frac{(-1)^{n+3}}{n!} = (-1)^3 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = - e^{-1}$.Evaluate second sum: let $m = k-1$. Then $k+1 = m+2$, sum becomes $2 \sum_{m=0}^{\infty} \frac{(-1)^{m+2}}{m!} = 2 \cdot (-1)^2 \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} = 2 e^{-1}$.So $S = -e^{-1} + 2e^{-1} = e^{-1}$.Answer:$S = \frac{1}{e}$ which corresponds to option D.