Comparison and Limit Comparison Tests — Question 4

*Question 4 Consider $\displaystyle\sum_{n=1}^{\infty}\frac1{\sqrt{n^2+1}}$.

1. Select a benchmark using dominant powers.

2. Perform limit comparison.

3. Prove divergence again using a direct lower bound.

*Question 4 – Solution

Since $\sqrt{n^2+1}$ behaves like $n$, choose $b_n=1/n$.

$$\lim_{n\to\infty}\frac{1/\sqrt{n^2+1}}{1/n} =\lim_{n\to\infty}\frac1{\sqrt{1+1/n^2}}=1.$$ The finite positive ratio and divergence of $\sum1/n$ imply divergence.

Since $n^2+1\le2n^2$ for $n\ge1$, $$\frac1{\sqrt{n^2+1}}\ge\frac1{\sqrt2\,n}.$$ The smaller comparison series diverges, independently confirming the verdict.