Comparison and Limit Comparison Tests — Question 2

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

1. Use dominant powers to select a harmonic benchmark.

2. Compute the Limit Comparison Test ratio.

3. Classify the series and give an independent direct lower comparison.

*Question 2 – Solution

The leading-term quotient is $2n/n^2=2/n$, so choose $b_n=1/n$. All terms are positive.

$$L=\lim_{n\to\infty}\frac{(2n+1)/(n^2+1)}{1/n} =\lim_{n\to\infty}\frac{2n^2+n}{n^2+1}=2.$$ Because $0<L<\infty$ and the harmonic series diverges, the given series diverges.

For $n\ge1$, $n^2+1\le2n^2$ and $2n+1\ge2n$, so $$\frac{2n+1}{n^2+1}\ge\frac{2n}{2n^2}=\frac1n.$$ A series larger than the divergent harmonic series must diverge.