Question

Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{\arctan n}{n^{2}+1} $$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{\arctan n}{n^{2}+1} \) is convergent.

Step-by-step solution

Verify the conditions of the Integral Test

Define the function \( f(x) = \frac{\arctan x}{x^2+1} \). The integral test applies when \( f(x) \) is positive, continuous, and decreasing when \( x >1 \). Given that \(\arctan x \) increases monotonically from \(0\) to \( \frac{\pi}{2} \) for \( x \geq 0 \) and \( x^2+1 > 1 \), \( f(x) \) is a positive and decreasing function when \( x > 1 \). By the nature of the \(\arctan x \) function, \( f(x) \) is continuous. Thus, \( f(x) \) meets the conditions for the Integral Test.

Apply the Integral Test

By the Integral Test, the convergence of the series is equivalent to the convergence of the integral \( \int_{1}^{\infty} \frac{\arctan x}{x^{2}+1} dx \). We can solve this definite integral by using a standard substitution method. Let \( u = \arctan x \) thus \( du = \frac{1}{1 + x^2} dx \) and our integral simplifies to \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} u du \). Evaluating this gives \( \frac{\pi^2}{16} - \frac{\pi^2}{32} = \frac{\pi^2}{32} \).

Conclusion

Because the integral from step 2 is finite, we can conclude, by the Integral Test, that the original series \( \sum_{n=1}^{\infty} \frac{\arctan n}{n^{2}+1} \) is convergent.