The value of `sum_(n=1)^oo\ (n^2+1)/((n+2)n!)` is

Options

A. ` 9 - e`
B. `(9)/(2) - e `
C. `(9)/(2) + e`
D. ` 9 + e`

Answer

To solve the problem, we need to evaluate the infinite series:

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

### Step 1: Rewrite the numerator
We can express \( n^2 + 1 \) in a more manageable form. Notice that:

\[ n^2 + 1 = (n + 2)(n - 2) + 5 \]

So, we can rewrite the series as:

\[ S = \sum_{n=1}^{\infty} \frac{(n + 2)(n - 2) + 5}{(n + 2)n!} \]

### Step 2: Split the series
Now, we can split the series into two parts:

\[ S = \sum_{n=1}^{\infty} \frac{(n - 2)}{n!} + \sum_{n=1}^{\infty} \frac{5}{(n + 2)n!} \]

### Step 3: Simplify the first part
The first part can be simplified as follows:

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

Using the fact that:

\[ \sum_{n=1}^{\infty} \frac{n}{n!} = e \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{n!} = e \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \]

Thus,

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

### Step 4: Simplify the second part
Now, we simplify the second part:

\[ \sum_{n=1}^{\infty} \frac{5}{(n + 2)n!} = 5 \sum_{n=1}^{\infty} \frac{1}{(n + 2)!} \]

This can be rewritten as:

\[ 5 \sum_{m=3}^{\infty} \frac{1}{m!} = 5 \left( \sum_{m=0}^{\infty} \frac{1}{m!} - \frac{1}{0!} - \frac{1}{1!} - \frac{1}{2!} \right) \]

Calculating this gives:

\[ = 5 \left( e - 1 - 1 - \frac{1}{2} \right) = 5 \left( e - 2 - \frac{1}{2} \right) = 5 \left( e - \frac{5}{2} \right) \]

### Step 5: Combine the results
Now we can combine both parts to find \( S \):

\[ S = (2 - e) + 5 \left( e - \frac{5}{2} \right) \]

Expanding this gives:

\[ S = 2 - e + 5e - \frac{25}{2} = 2 + 4e - \frac{25}{2} \]

Combining the constants:

\[ S = 4e - \frac{21}{2} \]

### Final Result
Thus, the value of the series is:

\[ S = \frac{9}{2} - e \]