Question

The value of the infinite series \[ \sum_{n=1}^{\infty} n \left( \frac{3}{4} \right)^{2n-1} \] is equal to ............. (rounded off to two decimal places).

Hint:

For series involving powers of a common ratio, use the standard formulas for geometric series to simplify the summation.

Answer 1

Step 1: Recognize the form of the series.
The given series is of the form: \[ \sum_{n=1}^{\infty} n \cdot r^{2n-1}, \] which is a geometric series with a common ratio \( r = \frac{3}{4} \). Step 2: Use the formula for the sum of such a series.
The sum of the series \( \sum_{n=1}^{\infty} n r^{n-1} \) is given by: \[ S = \frac{a}{(1 - r)^2}, \] where \( a \) is the first term and \( r \) is the common ratio. Step 3: Apply the formula.
For the given series, we apply the formula for the geometric series and find: \[ S = \frac{1}{\left(1 - \frac{3}{4}\right)^2} = \frac{1}{\left(\frac{1}{4}\right)^2} = 16. \] Final Answer: \[ \boxed{16}. \]