Question

use the direct comparison test to determine whether the following series converges or diverges.
sum _ { n = 1 } ^ { infty } \frac { sin ^ { 2 } n } { n ^ { 7 / 2 } }
choose the correct choice below.
○ a. the direct comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 7 / 2 } } ) shows that the series diverges.
○ b. the direct comparison test with ( sum _ { n = 1 } ^ { infty } sin ^ { 2 } n ) shows that the series converges.
○ c. the direct comparison test with ( sum _ { n = 1 } ^ { infty } sin ^ { 2 } n ) shows that the series diverges.
○ d. the direct comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 7 / 2 } } ) shows that the series converges.

Explanation:

Step1: Analyze the range of \(\sin^{2}n\)

Since \(- 1\leqslant\sin n\leqslant1\), then \(0\leqslant\sin^{2}n\leqslant1\). So \(\frac{\sin^{2}n}{n^{7/2}}\leqslant\frac{1}{n^{7/2}}\) for \(n = 1,2,\cdots\).

Step2: Recall the \(p -\)series test

The \(p -\)series \(\sum_{n = 1}^{\infty}\frac{1}{n^{p}}\) converges if \(p>1\) and diverges if \(p\leqslant1\). For the series \(\sum_{n = 1}^{\infty}\frac{1}{n^{7/2}}\), here \(p=\frac{7}{2}>1\), so \(\sum_{n = 1}^{\infty}\frac{1}{n^{7/2}}\) converges.

Step3: Apply the Direct Comparison Test

By the Direct Comparison Test, if \(0\leqslant a_{n}\leqslant b_{n}\) for all \(n\geqslant N\) (in this case \(N = 1\)) and \(\sum_{n = 1}^{\infty}b_{n}\) converges, then \(\sum_{n = 1}^{\infty}a_{n}\) converges. Let \(a_{n}=\frac{\sin^{2}n}{n^{7/2}}\) and \(b_{n}=\frac{1}{n^{7/2}}\).

Answer:

D. The Direct Comparison Test with \(\sum_{n = 1}^{\infty}\frac{1}{n^{7/2}}\) shows that the series converges.