The sum of the series `sum_(n=0)^(oo) (n^(2) - n +1)/(n!)` is

Options

A. e
B. `(3)/(2)e`
C. 2e
D. 3e

Answer

To find the sum of the series

\[ S = \sum_{n=0}^{\infty} \frac{n^2 - n + 1}{n!}, \]

we can break this series into three separate sums:

\[ S = \sum_{n=0}^{\infty} \frac{n^2}{n!} - \sum_{n=0}^{\infty} \frac{n}{n!} + \sum_{n=0}^{\infty} \frac{1}{n!}. \]

### Step 1: Evaluate \(\sum_{n=0}^{\infty} \frac{1}{n!}\)

This is the Taylor series expansion for \(e^x\) at \(x = 1\):

\[ \sum_{n=0}^{\infty} \frac{1}{n!} = e. \]

### Step 2: Evaluate \(\sum_{n=0}^{\infty} \frac{n}{n!}\)

We can rewrite \(\frac{n}{n!}\) as \(\frac{1}{(n-1)!}\) for \(n \geq 1\):

\[ \sum_{n=0}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e, \]

where we let \(m = n - 1\).

### Step 3: Evaluate \(\sum_{n=0}^{\infty} \frac{n^2}{n!}\)

We can use the identity \(n^2 = n(n-1) + n\):

\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = \sum_{n=0}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=0}^{\infty} \frac{n}{n!}. \]

The first term can be rewritten as:

\[ \sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e, \]

where we let \(m = n - 2\).

Thus, we have:

\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = e + e = 2e. \]

### Step 4: Combine all parts

Now we can combine all the parts:

\[ S = \sum_{n=0}^{\infty} \frac{n^2}{n!} - \sum_{n=0}^{\infty} \frac{n}{n!} + \sum_{n=0}^{\infty} \frac{1}{n!} = 2e - e + e = 2e. \]

Thus, the sum of the series is

\[ \boxed{2e}. \]