Limit comparison test

Statement

Let $(a_n)$ and $(b_n)$ be real sequences with $a_n \ge 0$ and $b_n > 0$ for all $n$. Assume that $a_n / b_n \to c$ as $n \to \infty$, for a real number $c$ with $0 < c < \infty$. Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Remarks

For large $n$ we have $\frac{c}{2} b_n \le a_n \le 2c\, b_n$. The comparison test then gives both directions. For example, $\frac{n}{n^3 + 1} / \frac{1}{n^2} \to 1$, so $\sum n/(n^3 + 1)$ converges, like $\sum 1/n^2$. If $c = 0$, convergence of $\sum b_n$ still implies convergence of $\sum a_n$. The converse can fail.

Explanations

L1 human

If two series of positive terms have terms of the same size in the long run, they behave the same way. So you can replace a complicated term by a simple one with the same leading behaviour.