`-(2)/(3)-(1)/(2)((4)/(9))-(1)/(3)((8)/(27))-(1)/(4)((16)/(81))-…..oo=`

Options

A. `-log 3
B. `-log_(e )2`
C. `log_(e )2`
D. `log_(e ) 3`

Answer

To solve the series

\[ S = -\frac{2}{3} - \frac{1}{2} \left(\frac{4}{9}\right) - \frac{1}{3} \left(\frac{8}{27}\right) - \frac{1}{4} \left(\frac{16}{81}\right) - \ldots \]

we can rewrite the series in a more manageable form.

### Step 1: Identify the pattern in the series

The series can be expressed as:

\[ S = -\frac{2}{3} - \frac{1}{2} \cdot \left(\frac{2^2}{3^2}\right) - \frac{1}{3} \cdot \left(\frac{2^3}{3^3}\right) - \frac{1}{4} \cdot \left(\frac{2^4}{3^4}\right) - \ldots \]

This can be generalized as:

\[ S = -\sum_{n=1}^{\infty} \frac{1}{n+1} \left(\frac{2^n}{3^n}\right) \]

### Step 2: Factor out the common term

We can factor out \(-\frac{1}{3}\):

\[ S = -\frac{1}{3} \sum_{n=1}^{\infty} \frac{2^n}{n \cdot 3^n} \]

### Step 3: Recognize the series as a logarithmic series

The series \(\sum_{n=1}^{\infty} \frac{x^n}{n}\) converges to \(-\log(1-x)\) for \(|x| < 1\). In our case, we can relate it to:

\[ \sum_{n=1}^{\infty} \frac{(2/3)^n}{n} = -\log(1 - \frac{2}{3}) = -\log(\frac{1}{3}) = \log(3) \]

### Step 4: Substitute back into the series

Thus, we have:

\[ S = -\frac{1}{3} \cdot \log(3) \]

### Step 5: Final result

The final result is:

\[ S = -\log(3) \]

### Summary

The series converges to \(-\log(3)\).