Find `(1)/(3)+(1)/(2.3^(2))+(1)/(3.3^(3))+(1)/(4.3^(4))+….oo=?`

Options

A. `log_(e )2-log_(e )3`
B. `log_(e )3-log_(e )2`
C. `log_(e )6`
D. `log_(e )3-log_(e )4`

Answer

To find the value of the series

\[ S = \frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \ldots \]

we can express the series in a more manageable form.

### Step 1: Rewrite the series
We can rewrite the series as follows:

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

### Step 2: Relate to the logarithmic series
We know that the series

\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) \]

for \(|x| < 1\). By differentiating both sides with respect to \(x\), we get:

\[ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x} \]

Multiplying both sides by \(x\):

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

### Step 3: Find the series we need
Now, we can relate our series to the logarithmic series. We have:

\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) \]

If we differentiate this with respect to \(x\):

\[ \sum_{n=1}^{\infty} n x^{n-1} = -\frac{1}{1-x} \]

Now, if we multiply by \(x\):

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

### Step 4: Substitute \(x = \frac{1}{3}\)
Now, substituting \(x = \frac{1}{3}\):

\[ \sum_{n=1}^{\infty} n \left(\frac{1}{3}\right)^n = -\frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = -\frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = -\frac{\frac{1}{3}}{\frac{4}{9}} = -\frac{3}{4} \]

### Step 5: Relate back to our series
Now, we have:

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

Using the property of logarithms:

\[ -\ln\left(\frac{2}{3}\right) = \ln\left(\frac{3}{2}\right) \]

### Final Result
Thus, we find that:

\[ S = \ln\left(\frac{3}{2}\right) \]