Find the sum of \(1-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-...\)
The given series is an alternating geometric series, with first term 1 and common ratio -1/2. The sum of an alternating geometric series is equal to the first term divided by 1 + the absolute value of the common ratio, so the sum of the given series is
1 / (1 + (-1/2)) = 1 / (3/2) = 2/3.
Therefore, the answer is 2/3.
