Q. 34 - Calculus

Question

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

      $\sum _{k=1}^{\mathrm{\infty }} \frac{{\tan }^{-1}k}{1+k^2}$

  

Step-by-Step Solution

Answer

Ans:  The series $\sum _{k=1}^{\mathrm{\infty }} \frac{{\tan }^{-1}k}{1+k^2}$ is convergent.

Step 1. Given information.

given,

    $\sum _{k=1}^{\mathrm{\infty }} \frac{{\tan }^{-1}k}{1+k^2}$

Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series.

Consider function $f\left(x\right)=\frac{{\tan }^{-1}x}{1+x^2}$.

The function $f\left(x\right)=\frac{{\tan }^{-1}x}{1+x^2}$ is continuous, decreasing, with positive terms. Therefore, all the conditions of the integral test are fulfilled. So, the integral test is applicable.

Step 3. Consider the integral ∫ x = 1 ∞   f ( x ) d x = ∫ x = 1 ∞   tan − 1 ⁡ x 1 + x 2 d x .

Therefore,

     $\begin{array}{r}{\int }_{x=1}^{\mathrm{\infty }} f\left(x\right)dx=\underset{k\to \mathrm{\infty }}{lim} {\int }_{x=1}^k \frac{{\tan }^{-1}x}{1+x^2}dx \\ =\underset{k\to \mathrm{\infty }}{lim} {\int }_{w=\frac{\pi }{4}}^{{\mathrm{man}}^{-1}k} udu\left(\text{ Put }{\tan }^{-1}x=u\Rightarrow \frac{1}{1+x^2}dx=du\right) \\ =\underset{k\to \mathrm{\infty }}{lim} {\left[\frac{u^2}{2}\right]}_{\frac{\pi }{4}}^{{\tan }^{-1}k} \text{ (Integrating) }\\ =\frac{1}{2}\underset{k\to \mathrm{\infty }}{lim} \left[{\left({\tan }^{-1}k\right)}^2-{\left(\frac{\pi }{4}\right)}^2\right] (\text{ Substitution) }\\ =\frac{1}{2}\left[{\left(\frac{\pi }{2}\right)}^2-{\left(\frac{\pi }{4}\right)}^2\right] \\ =\frac{1}{2}\left[\frac{{\pi }^2}{4}-\frac{{\pi }^2}{16}\right] \\ =\frac{1}{2}\left[\frac{3{\pi }^2}{16}\right] \\ =\frac{3{\pi }^2}{32} \text{ (Simplify) }\end{array}$

Step 4. Thus, the value of the integral is ∫ x = 1 ∞   tan − 1 ⁡ x 1 + x 2 d x = 3 π 2 32

The integral converges. Therefore, the series $\sum _{k=1}^{\mathrm{\infty }} \frac{{\tan }^{-1}k}{1+k^2}$ is convergent.

Hence, by integral test, the series $\sum _{k=1}^{\mathrm{\infty }} \frac{{\tan }^{-1}k}{1+k^2}$ is convergent and converges to $\frac{3{\mathrm{\pi }}^2}{32}$.