What is the value of this infinite sum Consider the series
$$\sum_{n=1}^{\infty}\frac{n^2-n+1}{n!}$$
From ratio test it is clear that this series is covergent. What is  its value  ?

We have $$\sum_{n=1}^{\infty} \frac {n^2-n+1}{n!} =\sum_{n=1}^{\infty} \frac {n^2}{n!} + \sum_{n=1}^{\infty}\frac {-n}{n!} +\sum_{n=1}^{\infty} \frac {1}{n!} = (2e)+ (- e) + (e-1) = 2e-1$$ Hope it helps.
