determine if the series converges or diverges. use any method, and give a reason for your answer\n\n sum _ {…

determine if the series converges or diverges. use any method, and give a reason for your answer\n\n sum _ { n = 1 } ^ { infty } \frac { sqrt { n } } { n ^ { 2 } + 5 } \n\nchoose the correct answer below\n\noa because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } geq \frac { 1 } { n ^ { 0.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 0.5 } } ) diverges, the series diverges by the direct comparison test\n\nob because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } geq \frac { 1 } { n ^ { 1.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 1.5 } } ) diverges, the series diverges by the direct comparison test\n\noc because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } leq \frac { 1 } { n ^ { 0.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 0.5 } } ) converges, the series converges by the direct comparison test\n\nod because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } leq \frac { 1 } { n ^ { 1.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 1.5 } } ) converges, the series converges by the direct comparison test

determine if the series converges or diverges. use any method, and give a reason for your answer\n\n sum _ { n = 1 } ^ { infty } \frac { sqrt { n } } { n ^ { 2 } + 5 } \n\nchoose the correct answer below\n\noa because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } geq \frac { 1 } { n ^ { 0.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 0.5 } } ) diverges, the series diverges by the direct comparison test\n\nob because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } geq \frac { 1 } { n ^ { 1.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 1.5 } } ) diverges, the series diverges by the direct comparison test\n\noc because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } leq \frac { 1 } { n ^ { 0.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 0.5 } } ) converges, the series converges by the direct comparison test\n\nod because ( \frac { sqrt { n } } { n ^ { 2 } + 5 } leq \frac { 1 } { n ^ { 1.5 } } ) and ( sum _ { n = 1 } ^ { infty } \frac { 1 } { n ^ { 1.5 } } ) converges, the series converges by the direct comparison test

Answer

Explanation:

Step1: Simplify the general term

We have (a_{n}=\frac{\sqrt{n}}{n^{2}+5}=\frac{n^{\frac{1}{2}}}{n^{2}+5}). For (n\geq1), (n^{2}+5\geq n^{2}), so (\frac{n^{\frac{1}{2}}}{n^{2}+5}\leq\frac{n^{\frac{1}{2}}}{n^{2}}=\frac{1}{n^{2 - \frac{1}{2}}}=\frac{1}{n^{1.5}}).

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\leq1). For the series (\sum_{n = 1}^{\infty}\frac{1}{n^{1.5}}), since (p = 1.5>1), the series (\sum_{n = 1}^{\infty}\frac{1}{n^{1.5}}) converges.

Step3: Apply the Direct Comparison Test

By the Direct Comparison Test, if (0\leq a_{n}\leq b_{n}) for all (n\geq N) (in this case (N = 1)) and (\sum_{n = 1}^{\infty}b_{n}) converges, then (\sum_{n = 1}^{\infty}a_{n}) converges. Here (a_{n}=\frac{\sqrt{n}}{n^{2}+5}) and (b_{n}=\frac{1}{n^{1.5}})

Answer:

D. Because (\frac{\sqrt{n}}{n^{2}+5}\leq\frac{1}{n^{1.5}}) and (\sum_{n = 1}^{\infty}\frac{1}{n^{1.5}}) converges, the series converges by the Direct Comparison Test.