If `agt0`, then `sum_(n=1)^(oo) ((a)/(a+1))^(n)` equals

Options

A. `(a+1)/(2a+1)`
B. `(a)/(2a+1)`
C. a+1
D. a

Explanation

To solve the problem, we need to find the sum of the infinite series given by:

\[ \sum_{n=1}^{\infty} \left( \frac{a}{a+1} \right)^{n} \]

where \( a > 0 \).

### Step-by-Step Solution:

1. **Identify the Series**:
The series can be rewritten as:
\[ \frac{a}{a+1} + \left( \frac{a}{a+1} \right)^{2} + \left( \frac{a}{a+1} \right)^{3} + \ldots \]
This is a geometric series where the first term \( a_1 = \frac{a}{a+1} \) and the common ratio \( r = \frac{a}{a+1} \).

2. **Check the Condition for Convergence**:
For a geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ \left| \frac{a}{a+1} \right| < 1 \]
Since \( a > 0 \), it follows that \( a + 1 > a \) and thus \( \frac{a}{a+1} < 1 \). Therefore, the series converges.

3. **Apply the Formula for the Sum of a Geometric Series**:
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a_1}{1 - r} \]
Substituting the values we have:
\[ S = \frac{\frac{a}{a+1}}{1 - \frac{a}{a+1}} \]

4. **Simplify the Denominator**:
To simplify \( 1 - \frac{a}{a+1} \):
\[ 1 - \frac{a}{a+1} = \frac{(a+1) - a}{a+1} = \frac{1}{a+1} \]

5. **Substitute Back into the Sum Formula**:
Now substituting back into the sum formula:
\[ S = \frac{\frac{a}{a+1}}{\frac{1}{a+1}} = a \]

6. **Final Result**:
Therefore, the sum of the series is:
\[ \sum_{n=1}^{\infty} \left( \frac{a}{a+1} \right)^{n} = a \]