Q. 47 - Calculus

Question

Use any convergence test from Sections 7.4–7.6 to determine whether the series in Exercises 41–59 converge or diverge. Explain why each series that meets the hypotheses of the test you select does so. 

$\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}$

Step-by-Step Solution

Answer

The given series converges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}.$

Step 2. Determine whether the series converges or diverges.

To determine whether the series converges or diverges we will use the comparison test since the series has non-negative terms that meet the hypothesis of the test.

Let $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}\mathrm{and} \sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2}.$

So, $0\le \frac{\sqrt{k}}{k^2+3}\le \frac{\sqrt{k}}{k^2}.$

We can write 

$$\begin{gathered} \sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2} \mathrm{as}: \\ \sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{{\mathrm{k}}^{{\displaystyle \frac{1}{2}}}}{{\mathrm{k}}^2} \\ \sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{1}{{\mathrm{k}}^{2-{\displaystyle \frac{1}{2}}}} \\ \sum _{\mathrm{k}=1}^{\infty }{\mathrm{b}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\infty }\frac{1}{{\mathrm{k}}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Now, $\sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{3}{2}}}} \mathrm{is} \mathrm{of} \mathrm{the} \mathrm{form} \sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^p}.$

If p > 1  then it is converges or if p < 1 then it is diverges.

Here, $p=\frac{3}{2}>1$ thus, $\sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2}$ converges.

By the comparison test, $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}$ also converges.

Thus, the given series converges.