The discussion focuses on the limit comparison test for the series sum of \( \frac{1}{\sqrt{n^2+1}} \) and its comparison with \( \frac{1}{2n} \). The user confirms that the inequality \( \frac{1}{\sqrt{n^2+1}} > \frac{1}{2n} \) holds true for \( n > \sqrt{\frac{1}{3}} \), which is valid for \( n \geq 1 \). This establishes that the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+1}} \) is divergent, as it can be compared to the divergent series \( \sum_{n=1}^{\infty} \frac{1}{2n} \).
PREREQUISITESStudents and educators in calculus, particularly those studying series convergence, as well as mathematicians interested in advanced series analysis techniques.
DrewD said:note that \sqrt{n^2+1} and 2n are positive here, so you don't have to worry about either one changing the direction of an inequality. Set up the inequality and do some algebra.