If `t_(n) = (Sigman)/(n!)`, then th sum of the infinite series `Sigma t_(n)` is equal to

Options

A. e
B. `e^(-1)`
C. `(3e)/(2)`
D. `(e)/(2)`

Answer

To solve the problem, we need to find the sum of the infinite series given by \( t_n = \frac{n}{n!} \). We will denote the sum of the series as \( S = \sum_{n=0}^{\infty} t_n \).

### Step-by-step Solution:

1. **Write the expression for \( t_n \)**:
\[ t_n = \frac{n}{n!} \]

2. **Rewrite \( t_n \)**:
We can rewrite \( t_n \) as:
\[ t_n = \frac{n}{n!} = \frac{1}{(n-1)!} \quad \text{for } n \geq 1 \]
Note that for \( n = 0 \), \( t_0 = 0 \).

3. **Express the sum \( S \)**:
The sum \( S \) can be expressed as:
\[ S = \sum_{n=1}^{\infty} \frac{n}{n!} \]

4. **Change the index of summation**:
We can change the index of summation by letting \( k = n - 1 \). Then \( n = k + 1 \) and the sum becomes:
\[ S = \sum_{k=0}^{\infty} \frac{k + 1}{(k + 1)!} = \sum_{k=0}^{\infty} \left( \frac{k}{(k + 1)!} + \frac{1}{(k + 1)!} \right) \]

5. **Split the sum**:
We can split this into two separate sums:
\[ S = \sum_{k=0}^{\infty} \frac{k}{(k + 1)!} + \sum_{k=0}^{\infty} \frac{1}{(k + 1)!} \]

6. **Simplify the first sum**:
The first sum can be simplified as follows:
\[ \sum_{k=0}^{\infty} \frac{k}{(k + 1)!} = \sum_{k=1}^{\infty} \frac{1}{k!} = e - 1 \]
(since \( \sum_{k=0}^{\infty} \frac{1}{k!} = e \)).

7. **Simplify the second sum**:
The second sum is:
\[ \sum_{k=0}^{\infty} \frac{1}{(k + 1)!} = \sum_{m=1}^{\infty} \frac{1}{m!} = e - 1 \]

8. **Combine the results**:
Now we can combine both results:
\[ S = (e - 1) + (e - 1) = 2e - 2 \]

9. **Final result**:
Therefore, the sum of the infinite series is:
\[ S = 2(e - 1) \]

### Conclusion:
Thus, the sum of the infinite series \( \sum_{n=0}^{\infty} t_n \) is equal to \( 2(e - 1) \).