The Comparison Test

Theorem 6.4.1 (The Comparison Test)   Suppose $\sum a_n$ and $\sum b_n$ are series with all $a_n$ and $b_n$ positive and $a_n\leq b_n$ for each $n$.

1. If $\sum b_n$ converges, then so does $\sum a_n$.
2. If $\sum a_n$ diverges, then so does $\sum b_n$.

Proof. [Proof Sketch] The condition of the theorem implies that for any $k$,

$$\sum_{n=1}^{k} a_n \leq \sum_{n=1}^k b_n,$$

from which each claim follows. $\qedsymbol$

Example 6.4.2   Consider the series $\sum_{n=1}^{\infty} \frac{7}{3n^2 + 2n}$. For each $n$ we have

$$\frac{7}{3n^2 + 2n} \leq \frac{7}{3} \cdot \frac{1}{n^2}.$$

Since $\sum_{n=1}^{\oo } \frac{1}{n^2}$ converges, Theorem 6.4.1 implies that $\sum_{n=1}^{\infty} \frac{7}{3n^2 + 2n}$ also converges.

Example 6.4.3   Consider the series $\sum_{n=1}^{\infty} \frac{\ln(n)}{n}$. It diverges since for each $n\geq 3$ we have

$$\frac{\ln(n)}{n} \geq \frac{1}{n},$$

and $\sum_{n=3}^{\infty} \frac{1}{n}$ diverges.