sum of the series `overset(n)underset(r=1)Sigma (r) /(r+1! ) is`

Options

A. `1-(1)/(n!)`
B. `1-(1)/((n+1)!)`
C. `2-(1)/((n+1)!)`
D. none of these

Answer

To find the sum of the series \( \sum_{r=1}^{n} \frac{r}{(r+1)!} \), we can follow these steps:

### Step 1: Rewrite the term
We start by rewriting the term \( \frac{r}{(r+1)!} \):
\[ \frac{r}{(r+1)!} = \frac{r+1 - 1}{(r+1)!} = \frac{r+1}{(r+1)!} - \frac{1}{(r+1)!} \]

### Step 2: Split the summation
Now we can split the summation into two parts:
\[ \sum_{r=1}^{n} \frac{r}{(r+1)!} = \sum_{r=1}^{n} \left( \frac{r+1}{(r+1)!} - \frac{1}{(r+1)!} \right) \]
This can be rewritten as:
\[ \sum_{r=1}^{n} \frac{r+1}{(r+1)!} - \sum_{r=1}^{n} \frac{1}{(r+1)!} \]

### Step 3: Simplify the first summation
The first summation can be simplified:
\[ \sum_{r=1}^{n} \frac{r+1}{(r+1)!} = \sum_{r=1}^{n} \frac{1}{r!} \]
This is because \( \frac{r+1}{(r+1)!} = \frac{1}{r!} \).

### Step 4: Simplify the second summation
The second summation can be rewritten by changing the index:
\[ \sum_{r=1}^{n} \frac{1}{(r+1)!} = \sum_{s=2}^{n+1} \frac{1}{s!} \]
where \( s = r + 1 \).

### Step 5: Combine the results
Now we can combine the results:
\[ \sum_{r=1}^{n} \frac{r}{(r+1)!} = \sum_{r=1}^{n} \frac{1}{r!} - \sum_{s=2}^{n+1} \frac{1}{s!} \]
This gives us:
\[ = \left( \sum_{r=1}^{n} \frac{1}{r!} \right) - \left( \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{(n+1)!} \right) \]

### Step 6: Final simplification
Notice that the terms from \( \frac{1}{2!} \) to \( \frac{1}{n!} \) will cancel out:
\[ = \frac{1}{1!} - \frac{1}{(n+1)!} \]
Thus, we have:
\[ = 1 - \frac{1}{(n+1)!} \]

### Final Answer
The sum of the series is:
\[ \sum_{r=1}^{n} \frac{r}{(r+1)!} = 1 - \frac{1}{(n+1)!} \]