How is $ \sum [(2/(n-1)!) + (1/n!)] = 2e + e - 1$?

Question

A step in the solution of a question I'm doing says that $ \sum [(2/(n-1)!) + (1/n!)] = 2e + e - 1$. How is this possible? Isn't e equal to $ 1/0! + 1/1! + 1/2!... $ till infinity? Also how does -1 come in the expression $2e+e-1$? I don't really understand how this is so. Would someone please explain this equation?

Answer 1

As $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$

$x=1\implies ?$

Now $\displaystyle\sum_{n=1}^\infty\left(\dfrac2{(n-1)!}+\dfrac1{n!}\right)$

$\displaystyle=2\sum_{n=1}^\infty\dfrac1{(n-1)!}+\sum_{n=1}^\infty\dfrac1{n!}$

$\displaystyle=2\sum_{m=0}^\infty\dfrac1{m!}+\sum_{n=0}^\infty\dfrac1{n!}-\dfrac1{0!}$

$=?$

Answer 2

HINT

Recall that

$$e^x=\sum_0^{\infty} \frac{x^n}{n!}\implies e=\sum_0^{\infty} \frac{1}{n!}$$