Comparison and Limit Comparison Tests

Comparison Tests for Series

By the end of this lesson, you will be able to apply the Direct Comparison Test and Limit Comparison Test to determine convergence or divergence.

Direct Comparison Test

Setup: suppose $0 \leq a_n \leq b_n$ for all $n \geq N$.

• If $\sum b_n$ converges → $\sum a_n$ converges (bounded by a convergent sum)
• If $\sum a_n$ diverges → $\sum b_n$ diverges (bigger than a divergent sum)

Key: both series must have positive terms, and the comparison must hold (at least eventually).

Example: Does $\sum \frac{1}{n^2+n}$ converge?

Compare with $\frac{1}{n^2}$: $\frac{1}{n^2+n} < \frac{1}{n^2}$ for all $n \geq 1$.

Since $\sum 1/n^2$ converges (p-series, $p=2$), so does $\sum \frac{1}{n^2+n}$. ✓

Limit Comparison Test

Setup: if $a_n > 0$, $b_n > 0$, and $\lim_{n\to\infty}\frac{a_n}{b_n} = L$ where $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Why: if $L$ is a finite positive number, then $a_n$ and $b_n$ "grow at the same rate" — one converges iff the other does.

Choosing $b_n$: use the dominant terms of $a_n$ for large $n$.

Example: Does $\sum \frac{3n+2}{5n^3-7}$ converge?

Dominant behavior: $\frac{3n}{5n^3} = \frac{3}{5n^2}$, so compare with $b_n = 1/n^2$.

$\lim\frac{a_n}{b_n} = \lim\frac{(3n+2)n^2}{5n^3-7} = \lim\frac{3n^3+2n^2}{5n^3-7} = \frac{3}{5}$ (finite, nonzero ✓)

Since $\sum 1/n^2$ converges, $\sum\frac{3n+2}{5n^3-7}$ converges. ✓

Summary

Direct Comparison: $0 \leq a_n \leq b_n$; if $\sum b_n$ converges → $\sum a_n$ converges.

Limit Comparison: $\lim a_n/b_n = L$ (finite, positive) → $\sum a_n$ and $\sum b_n$ have the same behavior.

Choosing the comparison: find the dominant terms of $a_n$ and use that as $b_n$ (typically a p-series or geometric).