How do you use the integral test to determine if #1/4+2/7+3/12+...+n/(n^2+3)+...# is convergent or divergent?

Question

Aurora Alvarado · Accepted Answer

The series #sum_(n=1)^(+oo) n/(n^2+3)# is divergent.

The integral tests states that a necessary and sufficient condition for: #sum_(n=1)^(+oo) a_n# to converge is that the integral: #int_1^(+oo) f(x)dx# is convergent, where #f(x)# is chosen such that #f(n) = a_n# If we pose: #f(x) = x/(x^2+3)# we can calculate: #int_1^(+oo) x/(x^2+3)dx= 1/2int_1^(+oo) (d(x^2+3))/(x^2+3)=[1/2ln(x^2+3)]_1^(+oo)# This integral is not convergent, and therefore the series is divergent.

Cameron Alexander · Answer

undefined To use the integral test to determine the convergence or divergence of the series $ \frac{1}{4} + \frac{2}{7} + \frac{3}{12} + \ldots + \frac{n}{n^2+3} + \ldots $, follow these steps:

1. Consider the function $ f(x) = \frac{x}{x^2 + 3} $.
2. Check if $ f(x) $ is continuous, positive, and decreasing for all $ x \geq 1 $.
3. Integrate $ f(x) $ from $ 1 $ to $ \infty $.
4. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Therefore, the convergence or divergence of the series is determined by the convergence or divergence of the integral $ \int_{1}^{\infty} \frac{x}{x^2 + 3} \, dx $.