Q. 36 - Calculus

Question

Use either the divergence test or the integral test to determine whether the series in Exercises 32–43 converge or diverge. Explain why the series meets the hypotheses of the test you select.

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

Step-by-Step Solution

Answer

The series is divergent.

Step 1. Given information

We have been given the series $\sum _{k=1}^{\infty }\frac{k}{k^2+3}$

We have to determine whether the series converge or diverge.

Step 2. Determine whether the series converge or diverge.

Consider function $f\left(x\right)=\frac{x}{3+x^2}$

The function is continuous, decreasing, with positive terms.

All the conditions of integral test are fulfilled.

So, integral test is applicable.

Consider the integral ${\int }_{x=1}^{\infty }f\left(x\right)dx={\int }_{x=1}^{\infty }\frac{x}{3+x^2}dx$

$$\begin{gathered} {\int }_{x=1}^{\infty }f\left(x\right)dx \\ =\underset{k\to \infty }{\lim }{\int }_{x=1}^k\frac{x}{3+x^2}dx \\ =\frac{1}{2}\underset{k\to \infty }{\lim }{\int }_{u=4}^{k^2+3}\frac{du}{u} (Put 3+x^2=u\Rightarrow 2xdx=du) \\ =\frac{1}{2}\underset{k\to \infty }{\lim }{\left[\ln \left|u\right|\right]}_4^{k^2+3} \\ =\frac{1}{2}\underset{k\to \infty }{\lim }\left[\ln \left|k^2+3\right|-\ln \left|4\right|\right] \\ =\infty \end{gathered}$$

The integral diverges.

So, the series is divergent.