which geometric series converges?\n$\frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}+cdots$\n$1+\frac{1}{2}+…

Explanation:

Step1: Recall convergence condition

A geometric series $\sum_{n = 1}^{\infty}a\cdot r^{n - 1}$ converges if $|r|\lt1$.

Step2: Analyze first series

For $\frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}+\cdots$, $a=\frac{1}{81}$ and $r = 3$. Since $|3|>1$, it diverges.

Step3: Analyze second series

For $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$, $a = 1$ and $r=\frac{1}{2}$. Since $\left|\frac{1}{2}\right|=\frac{1}{2}<1$, it converges.

Step4: Analyze third series

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

Step5: Analyze fourth series

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

Answer:

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