If in a series `t_n=n/((n+1)!)` then `sum_(n=1)^20 t_n` is equal to :

Options

A. `(20 !-1)/(20!)`
B. `(21!-1)/(21!)`
C. `(19!-1)/(20!)`
D. `(21!-1)/(20!)`

Answer

To solve the problem, we need to find the sum of the series defined by the term \( t_n = \frac{n}{(n+1)!} \) from \( n = 1 \) to \( n = 20 \).

### Step-by-Step Solution:

1. **Define the term**:
\[ t_n = \frac{n}{(n+1)!} \]

2. **Rewrite the term**:
We can rewrite \( t_n \) as:
\[ t_n = \frac{n + 1 - 1}{(n + 1)!} = \frac{n + 1}{(n + 1)!} - \frac{1}{(n + 1)!} \]
This simplifies to:
\[ t_n = \frac{1}{n!} - \frac{1}{(n + 1)!} \]

3. **Set up the summation**:
We need to find:
\[ S = \sum_{n=1}^{20} t_n = \sum_{n=1}^{20} \left( \frac{1}{n!} - \frac{1}{(n + 1)!} \right) \]

4. **Recognize the telescoping series**:
The series is telescoping, which means that most terms will cancel out:
\[ S = \left( \frac{1}{1!} - \frac{1}{2!} \right) + \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) + \ldots + \left( \frac{1}{20!} - \frac{1}{21!} \right) \]

5. **Simplify the sum**:
When we add these terms, we see that all intermediate terms cancel out:
\[ S = \frac{1}{1!} - \frac{1}{21!} \]
Thus, we have:
\[ S = 1 - \frac{1}{21!} \]

6. **Final result**:
Therefore, the sum \( \sum_{n=1}^{20} t_n \) is:
\[ S = 1 - \frac{1}{21!} \]

### Final Answer:
\[ \sum_{n=1}^{20} t_n = 1 - \frac{1}{21!} \]