which geometric series converges?\n∑_{n = 1}^∞ 1/6(4)^{n - 1}\n∑_{n = 1}^∞ 5(3/4)^{n - 1}\n∑_{n = 1}^∞…
which geometric series converges?\n∑_{n = 1}^∞ 1/6(4)^{n - 1}\n∑_{n = 1}^∞ 5(3/4)^{n - 1}\n∑_{n = 1}^∞ 3(7/5)^{n - 1}\n∑_{n = 1}^∞ 1/9(1)^{n - 1}
Answer
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 $\sum_{n = 1}^{\infty}\frac{1}{6}(4)^{n - 1}$, $r = 4$. Since $|4|=4>1$, it diverges.
Step3: Analyze second - series
For $\sum_{n = 1}^{\infty}5(\frac{3}{4})^{n - 1}$, $r=\frac{3}{4}$. Since $|\frac{3}{4}|=\frac{3}{4}<1$, it converges.
Step4: Analyze third - series
For $\sum_{n = 1}^{\infty}3(\frac{7}{5})^{n - 1}$, $r=\frac{7}{5}$. Since $|\frac{7}{5}|=\frac{7}{5}>1$, it diverges.
Step5: Analyze fourth - series
For $\sum_{n = 1}^{\infty}\frac{1}{9}(1)^{n - 1}$, $r = 1$. Since $|1| = 1$, it diverges.
Answer:
$\sum_{n = 1}^{\infty}5(\frac{3}{4})^{n - 1}$