The sum of `0.2+0.004+0. 00006+0. 0000008+...` to `oo` is

Options

A. `200/891`
B. `2000/9801`
C. `1000/9801`
D. `2180/9801`

Explanation

To find the sum of the series \(0.2 + 0.004 + 0.00006 + 0.0000008 + \ldots\) up to infinity, we can analyze the series and express it in a more manageable form.

### Step 1: Identify the terms of the series
The given series can be rewritten in a more structured way:
- The first term is \(0.2 = \frac{2}{10}\)
- The second term is \(0.004 = \frac{4}{1000} = \frac{4}{10^3}\)
- The third term is \(0.00006 = \frac{6}{100000} = \frac{6}{10^5}\)
- The fourth term is \(0.0000008 = \frac{8}{10000000} = \frac{8}{10^7}\)

So, we can express the series as:
\[ \frac{2}{10} + \frac{4}{10^3} + \frac{6}{10^5} + \frac{8}{10^7} + \ldots \]

### Step 2: Identify the pattern in the numerators
The numerators \(2, 4, 6, 8, \ldots\) form an arithmetic progression (AP) with:
- First term \(a = 2\)
- Common difference \(d = 2\)

The \(n\)-th term of this AP can be expressed as:
\[ a_n = 2n \]

### Step 3: Identify the pattern in the denominators
The denominators \(10, 10^3, 10^5, 10^7, \ldots\) can be expressed as:
\[ 10^{2n-1} \quad \text{for } n = 1, 2, 3, \ldots \]

### Step 4: Rewrite the series
Now we can rewrite the series as:
\[ \sum_{n=1}^{\infty} \frac{2n}{10^{2n-1}} \]

### Step 5: Factor out constants
We can factor out the constant \(2\):
\[ 2 \sum_{n=1}^{\infty} \frac{n}{10^{2n-1}} = 2 \cdot \frac{1}{10} \sum_{n=1}^{\infty} \frac{n}{10^{2n}} = \frac{2}{10} \sum_{n=1}^{\infty} \frac{n}{100^n} \]

### Step 6: Use the formula for the sum of \(n x^n\)
The sum \(\sum_{n=1}^{\infty} n x^n\) can be calculated using the formula:
\[ \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} \]
for \(|x| < 1\).

Here, \(x = \frac{1}{100}\):
\[ \sum_{n=1}^{\infty} n \left(\frac{1}{100}\right)^n = \frac{\frac{1}{100}}{\left(1 - \frac{1}{100}\right)^2} = \frac{\frac{1}{100}}{\left(\frac{99}{100}\right)^2} = \frac{1}{100} \cdot \frac{10000}{9801} = \frac{100}{9801} \]

### Step 7: Substitute back into the series
Now substituting back, we have:
\[ \frac{2}{10} \cdot \frac{100}{9801} = \frac{200}{98010} = \frac{20}{9801} \]

### Final Answer
Thus, the sum of the series \(0.2 + 0.004 + 0.00006 + 0.0000008 + \ldots\) is:
\[ \frac{20}{9801} \]