Let `S_(K)` be the sum of an infinte G.P. series whose first term is K and common ratio is `(K)/(K+1)(Kgt0)`. Then the value of `sum_(K=1)^(oo)((-1)^(K))/(S_(K))` is equal to

Options

A. `log_(e)4`
B. `log_(e)2-1`
C. `log_(e)2+1`
D. `1-log_(e)4`

Answer

To solve the problem, we need to find the value of the series:

\[ \sum_{K=1}^{\infty} \frac{(-1)^K}{S_K} \]

where \( S_K \) is the sum of an infinite geometric series with first term \( K \) and common ratio \( \frac{K}{K+1} \).

### Step 1: Find \( S_K \)

The formula for the sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = K \) and \( r = \frac{K}{K+1} \).

Substituting these values into the formula:

\[ S_K = \frac{K}{1 - \frac{K}{K+1}} = \frac{K}{\frac{K+1-K}{K+1}} = \frac{K}{\frac{1}{K+1}} = K(K+1) \]

### Step 2: Substitute \( S_K \) into the series

Now we substitute \( S_K \) back into the series:

\[ \sum_{K=1}^{\infty} \frac{(-1)^K}{S_K} = \sum_{K=1}^{\infty} \frac{(-1)^K}{K(K+1)} \]

### Step 3: Simplify the series using partial fractions

We can express \( \frac{1}{K(K+1)} \) using partial fractions:

\[ \frac{1}{K(K+1)} = \frac{1}{K} - \frac{1}{K+1} \]

Thus, we can rewrite the series:

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

### Step 4: Split the series

Now, we can split the series into two separate series:

\[ \sum_{K=1}^{\infty} \frac{(-1)^K}{K} - \sum_{K=1}^{\infty} \frac{(-1)^K}{K+1} \]

### Step 5: Adjust the second series

The second series can be rewritten by changing the index of summation:

\[ \sum_{K=1}^{\infty} \frac{(-1)^K}{K+1} = \sum_{J=2}^{\infty} \frac{(-1)^{J-1}}{J} = -\sum_{J=2}^{\infty} \frac{(-1)^J}{J} \]

### Step 6: Combine the series

Now we have:

\[ \sum_{K=1}^{\infty} \frac{(-1)^K}{K} + \sum_{J=2}^{\infty} \frac{(-1)^J}{J} \]

This simplifies to:

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

### Step 7: Recognize the series

The series \( \sum_{J=1}^{\infty} \frac{(-1)^J}{J} \) converges to \( -\ln(2) \). Therefore, we can write:

\[ \sum_{J=2}^{\infty} \frac{(-1)^J}{J} = -\ln(2) + 1 \]

### Step 8: Final calculation

Putting everything together, we have:

\[ -1 + 2(-\ln(2) + 1) = -1 - 2\ln(2) + 2 = 1 - 2\ln(2) \]

Thus, the final answer is:

\[ \boxed{1 - 2\ln(2)} \]