Question

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}-\frac{1}{3^{n}}\right) $$

Short answer

The sum of the inconvergent series is \( \frac{1}{2} \).

Step-by-step solution

Identify the Terms and Calculate the Sum for each Series

First we can recognize that the series is a sum of two geometric series. We separate them into two parts and get \( \sum_{n=0}^{\infty}\frac{1}{2^{n}} \) and \( \sum_{n=0}^{\infty}\frac{1}{3^{n}} \). For each series, we can use the sum formula for infinite geometric series \[ S = \frac{a}{1-r} \], where \( a \) is the first term of the series and \( r \) is the common ratio.

Calculate and Subtract the Sums

For the sum of the first geometric series, we get: \( S_1 = \frac{1}{1 - \frac{1}{2}} = 2 \). For the sum of the second geometric series, we get: \( S_2 = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \). Finally, the result of the whole series is obtained by subtracting the sum of the second geometric series from the sum of the first geometric series, \( S = S_1 - S_2 = 2 - \frac{3}{2} \).

Simplify the Result

This will yield the final result of the problem. Simplify the result to get: \( S = \frac{1}{2} \).