Question

How do you use the integral test to determine if, $\sum\limits_{n=1}^{\infty }{\dfrac{\arctan \left( n \right)}{{{n}^{2}}+1}}$, is convergent or divergent?

Answer 1

Integral test is a method to check whether the given series converges or diverges. If the given series converges, then the summation of the series is finite and we can calculate this value and if the series diverges, then the summation of the series approaches infinity. To check for convergence or divergence, we try to convert the given series into an integral of a continuous decreasing function and find its value.

Complete step by step solution:
Let us test the convergence or divergence of the given expression. We will write the sum of series as an integral and then calculate it. This can be done as follows:
$\Rightarrow I=\int\limits_{1}^{\infty }{\dfrac{\arctan x}{{{x}^{2}}+1}dx}$
Where, ‘I’ is the result of this integral. This equation could be written as:
$\Rightarrow I=\displaystyle \lim_{b \to \infty }\int\limits_{1}^{b}{\dfrac{\arctan x}{{{x}^{2}}+1}dx}$
Now, let us substitute $u\to \arctan x$. Then, we have:
$\Rightarrow du=\dfrac{1}{{{x}^{2}}+1}dx$
Substitution will also change the limits of the integral. So, our new integral could be written as:
$\begin{aligned} & \Rightarrow I=\displaystyle \lim_{b \to \infty }\int\limits_{\dfrac{\pi }{4}}^{\arctan b}{u.du} \\\ & \Rightarrow I=\displaystyle \lim_{b \to \infty }\left[ \dfrac{{{u}^{2}}}{2} \right]_{\dfrac{\pi }{4}}^{\arctan b} \\\ & \Rightarrow I=\displaystyle \lim_{b \to \infty }\dfrac{1}{2}\left[ {{\arctan }^{2}}b-\dfrac{{{\pi }^{2}}}{16} \right] \\\ \end{aligned}$
Here, the value of, $\displaystyle \lim_{b \to \infty }\left[ \arctan b \right]=\dfrac{\pi }{2}$. Using this in our above equation, the integral can be further simplified as:
$\begin{aligned} & \Rightarrow I=\dfrac{1}{2}\left[ {{\left( \dfrac{\pi }{2} \right)}^{2}}-\dfrac{{{\pi }^{2}}}{16} \right] \\\ & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{16} \right] \\\ & \Rightarrow I=\dfrac{1}{2}\times \dfrac{3{{\pi }^{2}}}{16} \\\ & \therefore I=\dfrac{3{{\pi }^{2}}}{32} \\\ \end{aligned}$
Since the integral converges to give a finite value, the series converges as well.

Hence, the integral test shows that the series, $\sum\limits_{n=1}^{\infty }{\dfrac{\arctan \left( n \right)}{{{n}^{2}}+1}}$, is convergent in nature.

Note:
We used the integral test to check whether the given series is convergent or divergent. To check this, we assumed a continuous function similar to the general term of the series and integrated upon this function within the given limits. As we got a finite value, we said the series was convergent. When converting the series into an integral, we can use integral formulas and tricks to solve this integral.