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Comparison Test

Definition

Let $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ be positive-term series, and $a_n \leq b_n$ (for sufficiently large $n$ ), then:

If $\sum_{n=1}^{\infty} b_n$ converges (convergence), then $\sum_{n=1}^{\infty} a_n$ converges (convergence)
If $\sum_{n=1}^{\infty} a_n$ diverges (divergence), then $\sum_{n=1}^{\infty} b_n$ diverges (divergence)

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\sum$	Greek letter	Sigma	Summation symbol, representing series
$\infty$	Mathematical symbol	Infinity	Represents infinite series, infinite number of terms

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ .

Solution: Since $\frac{1}{n^2 + 1} < \frac{1}{n^2}$ , and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ ),

Therefore, by the comparison test, $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges.

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$ .

Reference Answer

series convergencecomparison testp-series

Approach: Use the comparison test to compare with a known convergent series.

Steps:

Since $\frac{1}{n^2 + n} < \frac{1}{n^2}$ , and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ )
By the comparison test, $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$ converges

Answer: The series converges (convergence).

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$n$	Mathematical symbol	Number of terms	Number of terms in the series

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