Let `S_(k)`, be the sum of an infinite geometric series whose first term is kand common ratio is `(k)/(k+1)(kgt0)` . Then the value of `sum_(k=1)^(oo)(-1)^(k)/(S_(k))` is equal to

Options

A. `1-4 "log"_(e) sqrt(2)`
B. `"log"_(e) 2-1`
C. `1-"log"_(e)2`
D. `1-"log"_(e)4`

Answer

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

\[ \sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_k} \]

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

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

The sum \( S_k \) of an infinite geometric series can be calculated using the formula:

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

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 gives:

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

### Step 2: Substitute \( S_k \) into the sum

Now, we substitute \( S_k \) back into our original sum:

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

### Step 3: Simplify the term \( \frac{1}{k(k+1)} \)

We can simplify \( \frac{1}{k(k+1)} \) using partial fractions:

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

### Step 4: Rewrite the series

Now we can rewrite the sum:

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

This can be split into two separate sums:

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

### Step 5: Change the index of the second sum

For the second sum, we can change the index by letting \( j = k + 1 \):

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

### Step 6: Combine the sums

Now we can combine the two sums:

\[ \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} - \sum_{j=2}^{\infty} \frac{(-1)^{j-1}}{j} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} - \left( \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} - \frac{(-1)^{1}}{1} \right) \]

This simplifies to:

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

### Step 7: Evaluate the alternating series

The series \( \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} \) is known to converge to \( -\log(2) \). Thus, we have:

\[ -\log(2) + 1 \]

### Final Result

Therefore, the final value of the sum is:

\[ 1 - \log(2) \]