Q. 48 - Calculus

Question

Use any convergence test from this section or the previous section to determine whether the series in Exercises 31–48 converge or diverge. Explain how the series meets the hypotheses of the test you select.

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given series is the following.

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

Step 2. The Limit Comparison Test.

Consider a series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{3}{2}}}}$by taking the dominant term of numerator and denominator of $\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}.$

Find the value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}.$

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{\frac{\sqrt{k}}{k^2+3}}{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle k^{\frac{3}{2}}}}}} \\ \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{k^{\frac{1}{2}}{k}^{\frac{3}{2}}}{{\displaystyle k^2+3}} \\ \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{k^2}{{\displaystyle k^2+3}} \\ \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\frac{1}{1+{\displaystyle \frac{3}{k^2}}} \\ \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=1 \end{gathered}$$

Since $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{3}{2}}}}$is convergent by the p-series test so $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2+3}$is also convergent.