which geometric series converges?\n\\(\\frac{1}{81}+\\frac{1}{27}+\\frac{1}{9}+\\frac{1}{3}+\\cdots\\)\n\\(1…

Answer:

1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

Explanation:

Step1: Recall geometric - series convergence condition

A geometric series (\sum_{n = 1}^{\infty}a\cdot r^{n - 1}) converges if (|r|\lt1) and diverges if (|r|\geq1).

Step2: Analyze the first series

For the series (\frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}+\cdots), (a=\frac{1}{81}) and (r = 3). Since (|r|=3\gt1), it diverges.

Step3: Analyze the second series

For the series (1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots), (a = 1) and (r=\frac{1}{2}). Since (|r|=\frac{1}{2}\lt1), it converges.

Step4: Analyze the third series

For the series (\sum_{n = 1}^{\infty}7(-4)^{n - 1}), (a = 7) and (r=-4). Since (|r| = 4\gt1), it diverges.

Step5: Analyze the fourth series

For the series (\sum_{n = 1}^{\infty}\frac{1}{5}(2)^{n - 1}), (a=\frac{1}{5}) and (r = 2). Since (|r|=2\gt1), it diverges.