Convergence and Divergence of Series — Question 9

*Question 9 Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ converges or diverges.

1. Verify the nth-term condition.

2. Perform a limit comparison with the harmonic series.

3. Give a direct lower comparison that independently proves divergence.

*Question 9 – Solution

Dividing by $n^2$ shows $n/(n^2+1)=(1/n)/(1+1/n^2)\to0$, so the nth-term test is inconclusive. The dominant-power behavior suggests the harmonic benchmark $b_n=1/n$.

$$\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n} =\lim_{n\to\infty}\frac{n^2}{n^2+1}=1.$$ The limit equals $1$, which is finite and strictly positive. Since both series have positive terms and $\sum1/n$ diverges, the Limit Comparison Test proves that the given series diverges.

For $n\ge1$, $n^2+1\le2n^2$, so taking positive reciprocals and multiplying by $n$ gives $$\frac{n}{n^2+1}\ge\frac{n}{2n^2}=\frac1{2n}.$$ The smaller series $\sum1/(2n)$ diverges; therefore the larger given positive-term series also diverges. This direction of comparison is essential when proving divergence.