Wiskunde Multiple Choice Vraag

$\sum_{n=2}^{+\infty}\,\frac{n-1}{(n+1)!}=$ (e = 2,718281828459045. . .)	A. e − 1
B. e − 2
C. 3 − e
D. 2 −
E. 0

$\inline\\\displaystyle\sum_{n=2}^{+\infty}\,\frac{n-1}{(n+1)!}=\sum_{n=2}^{+\infty}\,\frac{n+1-2}{(n+1)!}=\sum_{n=2}^{+\infty}\,\frac{n+1}{(n+1)!}-2\sum_{n=2}^{+\infty}\,\frac{1}{(n+1)!}=\sum_{n=2}^{+\infty}\,\frac{1}{n!}\,-\,2\sum_{n=2}^{+\infty}\,\frac{1}{(n+1)!}\\=\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)-2\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots\right)$
Nu is $\textbf{e}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots=2+\frac{1}{2!}+\frac{1}{3!}+\cdots$
zodat (*) gelijk is aan (e − 2) − 2(e − 2 − 1op2

) = e − 2 − 2e + 4 + 1 = 3 − e ≈ 0,2817