How do we calculate this?
$$ \sum_{n=1}^\infty \frac{n-1}{n!}$$
From my readings on previous posts here and through Google, I found that:
$$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$
But then how do I manipulate that to help me find the answer? Through Wolfram Alpha, I know the answer is 1 but I don't know how to get to it. Thanks :)
On
Note: $${n - 1 \over n!} = {n \over n!} - {1 \over n!} = {n \over n\times (n-1)!} - {1 \over n!} = {1 \over (n-1)!} - {1 \over n!}$$
Write this as telescoping and you should see that many terms cancel out.
So $$\begin{align}\sum_{n=1}^\infty\left[{1 \over (n-1)!} - {1 \over n!}\right] &= (1 - 1) + \left(1 - {1 \over 2}\right) + \left({1 \over 2} - {1 \over 6}\right) + \left({1 \over 6} - {1 \over 24}\right) + \cdots \\ &=0 + 1 + \left(- {1 \over 2} + {1 \over 2}\right) + \left(-{1 \over 6} + {1 \over 6}\right) + \left(-{1 \over 24} + {1 \over 24}\right) + \cdots \\ & = 1 + 0 + 0 +\cdots \\ &= 1\end{align}$$
Break it up into two sums:
$$\sum_{i=1}^\infty\frac{n-1}{n!}=\sum_{i=1}^\infty\frac{n}{n!}-\sum_{n=1}^\infty\frac{1}{n!}.$$ Now note that $n/n!=1/(n-1)!.$