9 multiple choice 1 point which of the following series converge? i. ∑n = 1∞1n ii. ∑n = 1∞3nn! iii. ∑n =…

Answer:

E. II and III only

Explanation:

Step1: Analyze series I

The series $\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ is a p - series with $p=\frac{1}{2}$. Since $p=\frac{1}{2}<1$, by the p - series test $\sum_{n = 1}^{\infty}\frac{1}{n^{p}}$ diverges for $p\leq1$. So series I diverges.

Step2: Analyze series II

Use the ratio test. Let $a_{n}=\frac{3^{n}}{n!}$. Then $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{3^{n+1}}{(n + 1)!}}{\frac{3^{n}}{n!}}\right|=\lim_{n\rightarrow\infty}\left|\frac{3^{n+1}n!}{3^{n}(n + 1)!}\right|=\lim_{n\rightarrow\infty}\frac{3}{n+1}=0<1$. By the ratio test, $\sum_{n = 1}^{\infty}\frac{3^{n}}{n!}$ converges.

Step3: Analyze series III

The series $\sum_{n = 1}^{\infty}(\frac{e}{\pi})^{n}$ is a geometric series with common ratio $r=\frac{e}{\pi}$. Since $e\approx2.718$ and $\pi\approx3.14$, then $|r|=\frac{e}{\pi}<1$. By the geometric - series test $\sum_{n = 1}^{\infty}ar^{n}$ converges for $|r|<1$. So series III converges.