Mathematics

The key is to recognize that for large $n$, this series behaves like $\frac{n}{n^2} = \frac{1}{n}$, which diverges. Use the limit comparison test by comparing your series to $\frac{1}{n}$: compute $\lim_{n \to \infty} \frac{(n+3)/(n^2+1)}{1/n}$ and check whether it's a positive finite number. If so, both series share the same fate.

The key is to recognize that for large $n$, this series behaves like $\frac{n}{n^2} = \frac{1}{n}$, which diverges. Use the limit comparison test by comparing your series to $\frac{1}{n}$: compute $\lim_{n \to \infty} \frac{(n+3)/(n^2+1)}{1/n}$ and check whether it's a positive finite number. If so, both series share the same fate.