10. which of the following series converge? i. ∑(n = 1 to ∞) n/(13n + 4) ii. ∑(n = 0 to ∞) 5/3^(-n) iii. ∑(n…

Explanation:

Step1: Check series I using the limit - comparison test

Let (a_n=\frac{n}{13n + 4}) and (b_n=\frac{1}{13}). Then (\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{\frac{n}{13n + 4}}{\frac{1}{13}}=\lim_{n\rightarrow\infty}\frac{13n}{13n + 4}=\lim_{n\rightarrow\infty}\frac{13}{13+\frac{4}{n}} = 1). Since (\sum_{n = 1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{13}) diverges (it is a non - zero constant series), and (\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=1>0), by the limit - comparison test, (\sum_{n = 1}^{\infty}\frac{n}{13n + 4}) diverges.

Step2: Check series II using the geometric series formula

Rewrite the series (\sum_{n = 0}^{\infty}5\cdot3^{-n}=\sum_{n = 0}^{\infty}5\cdot(\frac{1}{3})^n). For a geometric series (\sum_{n=0}^{\infty}ar^n), where (a = 5) and (r=\frac{1}{3}). Since (|r|=\left|\frac{1}{3}\right|<1), by the geometric series test ((\sum_{n = 0}^{\infty}ar^n=\frac{a}{1 - r}) for (|r|<1)), the series (\sum_{n = 0}^{\infty}5\cdot3^{-n}) converges.

Step3: Check series III using the p - series test

Rewrite the series (\sum_{n = 1}^{\infty}\frac{4}{n\cdot\sqrt[3]{n}}=4\sum_{n = 1}^{\infty}\frac{1}{n^{1+\frac{1}{3}}}=4\sum_{n = 1}^{\infty}\frac{1}{n^{\frac{4}{3}}}). For a p - series (\sum_{n = 1}^{\infty}\frac{1}{n^p}), when (p=\frac{4}{3}>1), by the p - series test ((\sum_{n = 1}^{\infty}\frac{1}{n^p}) converges if (p > 1) and diverges if (p\leqslant1)), the series (\sum_{n = 1}^{\infty}\frac{4}{n\cdot\sqrt[3]{n}}) converges.

Answer:

E. II and III