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  ?
 A: It may be rewritten as
$$\frac{n^2-n+1}{n!}=\frac n{(n-1)!}-\frac1{(n-1)!}+\frac1{n!}$$
So the last two terms are telescoping and equal to $-1$.  The rest may swiftly be taken care of as well:
$$\frac n{(n-1)!}=\frac{n-1}{(n-1)!}+\frac1{(n-1)!}=\frac1{(n-2)!}+\frac1{(n-1)!}$$
where $\frac1{(-1)!}=0$, giving us

$$S=-1+2\sum_{n=0}^\infty\frac1{n!}=2e-1$$

A: Hint: Start with the series for $e^x$, and remember that $e^x$ doesn't change upon differentiation. So, upon successive differentiation, you have
$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$
$$e^x = \sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!}$$
$$e^x = \sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{n!}$$
Now put $x=1$ and note that you may have to supply some initial terms to recover your series by combining these series.
A: $\sum_{n=1}^{\infty} \frac{n^2-n+1}{n!}=\sum_{n=1}^{\infty} \Big( \frac{1}{(n-2)!}+\frac{1}{n!}\Big)= \sum_{n=2}^{\infty} \frac{1}{(n-2)!}+ \sum_{n=0}^{\infty} \frac{1}{n!}-1=2e-1$
A: Let $f(x)=\sum_{n=1}^\infty\frac{x^n}{n!}$. Then
$$ f(x)=e^x-1,f'(x)=\sum_{n=1}^\infty\frac{nx^{n-1}}{n!}=e^x,(xf'(x))'=\sum_{n=1}^\infty\frac{n^2x^{n-1}}{n!}=(x+1)e^x-1$$
and hence
$$ \sum_{n=1}^{\infty}\frac{n^2-n+1}{n!}=(x+1)e^x-1-e^x+e\bigg|_{x=1}=2e-1.$$
A: This is equal to
\begin{align*}
\sum_{n=1}^{\infty}\frac{n}{(n-1)!}-\sum_{n=0}^\infty\frac{1}{n!}+\sum_{n=1}^\infty\frac{1}{n!}&=\sum_{n=1}^{\infty}\frac{n}{(n-1)!}-1 =\sum_{n=0}^{\infty}\frac{n+1}{n!}-1\\\
&=\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}-1 \\\
&=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+e-1=2e-1.
\end{align*}
A: 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.
