Question

determine if the series converges or diverges. use any method, and give a reason for your answer

sum _ { n = 1 } ^ { infty } \frac { sqrt { n } } { n ^ { 2 } + 5 }

choose the correct answer below

oa 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

ob 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

oc 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

od 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

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.