Comparison Test for Improper Integrals — Question 6

Question 6

A student uses $$\frac1{\sqrt{x^2+1}}<\frac1x$$ to claim convergence on $[1,\infty)$. Explain the error and classify the integral.

Question 6 – Solution

Step 1: Identify the error. The comparison integral $$\int_1^\infty\frac{dx}{x}$$ diverges. Being smaller than a divergent function does not prove convergence; a smaller positive function may converge or diverge.

Step 2: Use limit comparison with $g(x)=1/x$. $$\begin{align*} L&=\lim_{x\to\infty} \frac{1/\sqrt{x^2+1}}{1/x}\\ &=\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}\\ &=\lim_{x\to\infty}\frac1{\sqrt{1+1/x^2}}=1. \end{align*}$$ Step 3: Apply limit comparison. Since $0<L<\infty$ and the harmonic integral diverges, the original integral also diverges. $$\boxed{\text{diverges by limit comparison}}$$