does the series below converge or diverge? give a reason for your answer. (when checking your answer…
does the series below converge or diverge? give a reason for your answer. (when checking your answer, remember there may be more than one way to determine the series convergence or divergence.)\n∑n = 2 to ∞ 7√n / ln 8n\ndoes the series converge or diverge? why or why not?\na. the series converges. this is revealed by the integral test.\nb. the series diverges. this is revealed by the nth - term test.\nc. the series diverges. this is revealed by rewriting the series as a geometric series with |r| > 1.\nd. the series converges. this is revealed by rewriting the series as a geometric series with |r| < 1.
Answer
Explanation:
Step1: Apply n - th term test
We find the limit as (n\rightarrow\infty) of (a_{n}=\frac{7\sqrt{n}}{\ln(8n)}). As (n\rightarrow\infty), we use L'Hopital's rule. Let (y = \frac{7\sqrt{n}}{\ln(8n)}), rewrite (\sqrt{n}=n^{\frac{1}{2}}). The derivative of the numerator (u = 7n^{\frac{1}{2}}) is (u'=\frac{7}{2\sqrt{n}}), and the derivative of the denominator (v=\ln(8n)) is (v'=\frac{1}{n}).
Step2: Calculate the limit after L'Hopital's rule
(\lim_{n\rightarrow\infty}\frac{7\sqrt{n}}{\ln(8n)}=\lim_{n\rightarrow\infty}\frac{\frac{7}{2\sqrt{n}}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{7n}{2\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{7}{2}\sqrt{n}=\infty). Since (\lim_{n\rightarrow\infty}a_{n}\neq0), by the n - th term test for divergence, the series (\sum_{n = 2}^{\infty}\frac{7\sqrt{n}}{\ln(8n)}) diverges.
Answer:
B. The series diverges. This is revealed by the nth - term test.