The value of `1/3+1/(3*3^(3))+1/(5*3^(5))+(1)/(7*3^(7))+. . .` is

Options

A. `1/2log_(e)2`
B. `log_(e)2`
C. `log_(e)3`
D. None of these

Answer

To solve the series \( S = \frac{1}{3} + \frac{1}{3 \cdot 3^3} + \frac{1}{5 \cdot 3^5} + \frac{1}{7 \cdot 3^7} + \ldots \), we can express it in a more manageable form and utilize properties of logarithms.

### Step 1: Rewrite the series
We can rewrite the series as:
\[ S = \sum_{n=1}^{\infty} \frac{1}{(2n-1) \cdot 3^{2n-1}} \]
This means we are summing over odd integers in the denominator while raising 3 to the power of the odd integers.

### Step 2: Relate to logarithmic series
We know that:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
and
\[ \log(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) \]
By subtracting these two series, we can isolate the odd powers of \( x \).

### Step 3: Combine logarithmic identities
We can express:
\[ \log(1+x) - \log(1-x) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]
This means:
\[ \log\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots\right) \]

### Step 4: Substitute \( x = \frac{1}{3} \)
Now, substituting \( x = \frac{1}{3} \):
\[ \log\left(\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\right) = \log\left(\frac{\frac{4}{3}}{\frac{2}{3}}\right) = \log(2) \]

### Step 5: Simplify the equation
This gives us:
\[ \log(2) = 2\left(\frac{1}{3} + \frac{1}{3^3 \cdot 3} + \frac{1}{5 \cdot 3^5} + \ldots\right) \]
Thus, we can express our series \( S \) as:
\[ S = \frac{1}{2} \log(2) \]

### Final Result
The value of the series \( S \) is:
\[ S = \frac{1}{2} \log(2) \]