Integral and Comparison Tests

Compare quantities with known behavior

For $0\le a_n\le b_n$ eventually, convergence of $\sum b_n$ forces convergence of $\sum a_n$ . Divergence of $\sum a_n$ forces divergence of $\sum b_n$ . A smaller divergent series or a larger convergent series is useful; the reversed comparisons usually say nothing.

Changing finitely many terms does not change convergence. We may start an inequality after any fixed index, while remembering that it can change the numerical sum.

Visual guide

VISUAL GUIDECompare positive terms with a known convergent series

1/n²: upper bound
1/(n² + 1): target

For n ≥ 1, 1/(n² + 1) ≤ 1/n². Since the upper comparison series converges, the smaller positive series also converges. The ordering of the dots is the inequality needed by the test.

Worked example: the integral test

If $f$ is positive, continuous and decreasing for $x\ge N$ , and $a_n=f(n)$ , then $\sum a_n$ and $\int_N^\infty f(x)dx$ either both converge or both diverge. Comparing rectangles with the area under the graph establishes the result.

Applying it to $f(x)=x^{-p}$ gives the $p$ -series rule: $\sum_{n=1}^\infty1/n^p$ converges exactly when $p>1$ . For $p>0$ use the integral test; for $p\le0$ terms already fail to approach zero. In particular, the harmonic series diverges.

For a decreasing positive convergent series, the remainder after term $N$ satisfies

\int_{N+1}^\infty f(x)dx\le R_N\le\int_N^\infty f(x)dx.

For $1/n^2$ , this brackets the remainder between $1/(N+1)$ and $1/N$ .

Worked example: limit comparison

For $a_n=(3n+1)/(n^3+2)$ and $b_n=1/n^2$ , the ratio $a_n/b_n\to3$ . A positive finite ratio means their tails differ only by bounded positive factors, so both series converge.

The condition $0<L<\infty$ is essential for the two-way limit comparison conclusion. A ratio tending to zero needs a suitable one-way argument; it does not mean the numerator series automatically converges.

Practice

Test $\sum1/(n^2+5)$ .
Test $\sum1/\sqrt n$ .
Bound the remainder after $100$ terms of $\sum1/n^2$ .

Show worked solutions

Since $0<1/(n^2+5)\le1/n^2$ , it converges.
It is a $p$ -series with $p=1/2\le1$ , so it diverges.
$1/101\le R_{100}\le1/100$ . The integral estimates bound the tail, not the total sum.