Comparison Tests

Comparison tests may mean either limit comparison tests or direct comparison tests. The limit comparison test is a method of testing for the convergence of an infinite series, while the direct comparison test is a way of deducing the convergence or divergence of an infinite series or an improper integral by comparison with other series or integral whose convergence properties are already known.

Limit Comparison Test

Statement: Suppose that we have two series, $\Sigma_n a_n$ and $\Sigma_n b_n$ , where $a_n$, $b_n$ are greater than or equal to $0$ for all $n$. If $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ with $0 < c < \infty$, then either both series converge or both series diverge.

Example: We want to determine if the series $\Sigma \frac{n+1}{2n^2}$ converges or diverges. For this we compare it with the series $\Sigma \frac{1}{n}$, which diverges. As $\lim_{n \to \infty} \frac{n+1}{2n^2} \frac{n}{1} = \frac{1}{2}$, we have that the original series also diverges.

Limit Convergence Test

The ratio between $\frac{n+1}{2n^2}$ and $\frac{1}{n}$ for $n \rightarrow ∞$ is $\frac{1}{2}$. Since the sum of the sequence $\frac{1}{n}$ $\left ( \text{i.e., }\sum {\frac{1}{n}}\right)$ diverges, the limit convergence test tells that the original series (with $\frac{n+1}{2n^2}$) also diverges.

Direct Comparison Test

The direct comparison test provides a way of deducing the convergence or divergence of an infinite series or an improper integral. In both cases, the test works by comparing the given series or integral to one whose convergence properties are known. In this atom, we will check the series case only.

For sequences $\{a_n\}$, $\{b_n\}$ with non-negative terms:

• If the infinite series $\sum b_n$ converges and $0 \le a_n \le b_n$ for all sufficiently large $n$ (that is, for all $n>N$ for some fixed value $N$), then the infinite series $\sum a_n$ also converges.
• If the infinite series $\sum b_n$ diverges and $a_n \ge b_n \ge 0$ for all sufficiently large $n$, then the infinite series $\sum a_n$ also diverges.

Example

The series $\Sigma \frac{1}{n^3 + 2n}$ converges because $\frac{1}{n^3 + 2n} < \frac{1}{n^3}$ for $n > 0$ and $\Sigma \frac{1}{n^3}$ converges.