Questions: Compute the value of the improper integral. (If the integral diverges to ∞, type oo; if it diverges to -∞, type -oo; if it diverges for some other reason, type DNE.) ∫1^∞ (5/(1+x^2)) dx= Use the value of the improper integral to determine whether or not the series ∑n=1^∞ (5/(1+n^2)) converges or diverges. Enter C if the series is convergent, D if the series is divergent, and ? if the Integral Test does not apply:

Solution Steps

To solve the given problem, we need to evaluate the improper integral and then use the result to determine the convergence of the series.

1. Improper Integral: The integral \(\int_{1}^{\infty} \frac{5}{1+x^{2}} dx\) is an improper integral because the upper limit is infinity. We can solve this by finding the antiderivative of the integrand and then evaluating the limit as the upper bound approaches infinity.

2. Series Convergence: The series \(\sum_{n=1}^{\infty} \frac{5}{1+n^{2}}\) can be analyzed using the Integral Test. If the integral converges, the series converges; if the integral diverges, the series diverges.

We need to compute the improper integral

\[ \int_{1}^{\infty} \frac{5}{1+x^{2}} \, dx. \]

To evaluate this, we find the antiderivative of the integrand. The antiderivative of \(\frac{5}{1+x^{2}}\) is

\[ 5 \tan^{-1}(x). \]

Thus, we can express the improper integral as

\[ \lim_{b \to \infty} \left[ 5 \tan^{-1}(x) \right]_{1}^{b} = \lim_{b \to \infty} \left( 5 \tan^{-1}(b) - 5 \tan^{-1}(1) \right). \]

As \(b\) approaches infinity, \(\tan^{-1}(b)\) approaches \(\frac{\pi}{2}\). Therefore, we have

\[ \lim_{b \to \infty} \left( 5 \cdot \frac{\pi}{2} - 5 \cdot \frac{\pi}{4} \right) = 5 \cdot \frac{\pi}{2} - \frac{5\pi}{4} = \frac{5\pi}{4}. \]

Next, we analyze the series

\[ \sum_{n=1}^{\infty} \frac{5}{1+n^{2}}. \]

Using the Integral Test, since the improper integral converges to \(\frac{5\pi}{4}\), we conclude that the series also converges.

Final Answer