§8.3 The Integral and Comparison Tests

1. Why we need these tests

When we meet an infinite series

$$\sum_{n=1}^{\infty} a_n ,$$

we really want to know only one thing first: Does the sum settle down to a finite value (converge) or blow up (diverge)? Most of the time the exact sum is impossible to compute, so we use tests that compare our series to something simpler or to an integral. The three big tools in this section are

1. Integral Test
2. Direct (ordinary) Comparison Test
3. Limit Comparison Test

All three assume the terms $a_n$ are positive (no negatives bouncing the sum around).

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2. Integral Test — “turn the sum into an area”

Conditions

• $f(x)$ is continuous, positive, and decreasing for $x\ge 1$.
• We set $a_n = f(n)$.

Statement

The series

$$\sum_{n=1}^{\infty} a_n$$

and the improper integral

$$\int_{1}^{\infty} f(x)\,dx$$

either both converge or both diverge.

Intuition (picture in words)

Place rectangles of width 1 whose heights match the graph of $f$ at integer points.

• Rectangles under the curve give an area ≤ the integral.
• Rectangles above the curve give an area ≥ the integral. If the integral area is finite, the stack of rectangles (your series) can’t exceed it too much, so it also converges.

Famous outcome – the $p$-series

$$\sum_{n=1}^{\infty} \frac{1}{n^{p}}$$

• Converges if $p>1$
• Diverges if $p\le 1$

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3. Direct (Ordinary) Comparison Test — “bigger‑smaller logic”

Suppose we have two positive‑term series $\sum a_n$ and $\sum b_n$.

How to choose $b_n$. Pick something you already understand—usually

• a $p$-series $\frac{1}{n^{p}}$, or
• a geometric series $ar^{n}$ with $|r|<1$.

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4. Limit Comparison Test — “compare long‑run ratios”

Sometimes the inequalities above are awkward. Instead we look at the ratio

$$c=\lim_{n\to\infty}\frac{a_n}{b_n},$$

where $c$ is a finite positive number. Then both series live or die together: if one converges, so does the other; if one diverges, so does the other.

A good strategy is to let $b_n$ be the dominant part of $a_n$ (e.g., keep the highest‑power term in the denominator).

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5. Standard “benchmarks” to remember

Keep these on speed‑dial; they will be your $b_n$ choices 90 % of the time.

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6. Step‑by‑step examples (mirroring your notes)

Example A 

$$\sum_{n=1}^{\infty}\frac{2}{3^{n}+4}$$

1. Pick comparator: $\displaystyle b_n=\frac{2}{3^{n}}$ (geometric).
2. Check $a_n \le b_n$. True because denominator of $a_n$ is bigger.
3. Geometric series converges ($r=\tfrac13$).
4. So our series converges by the Direct Comparison Test.

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Example B 

$$\sum_{n=1}^{\infty}\frac{n^{2}}{n^{3}+2}$$

1. Dominant behavior: $n^{2}/n^{3}\sim 1/n$.

2. Use $b_n=\frac1n$ (harmonic, divergent).

3. Compute limit:

   $$\lim_{n\to\infty}\frac{a_n}{b_n} =\lim_{n\to\infty}\frac{n^{2}/(n^{3}+2)}{1/n} =\lim_{n\to\infty}\frac{n^{3}}{n^{3}+2}=1>0.$$

4. By the Limit Comparison Test our series diverges (same fate as harmonic).

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Example C 

$$\sum_{n=2}^{\infty}\frac{\sqrt[3]{n^{2}+4}}{\sqrt{n^{3}(2n+5)}}$$

1. Simplify the dominant powers: numerator ≈ $n^{2/3}$; denominator ≈ $\sqrt{n^{3}\cdot 2n}=n^{2}$.
2. So $a_n\sim n^{2/3}/n^{2}=n^{-4/3}$.
3. Comparator $b_n=\dfrac{1}{n^{4/3}}$ is a $p$-series with $p=\tfrac{4}{3}>1$ → convergent.
4. Ratio limit is finite & positive → series converges by Limit Comparison.

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7. How to pick the right test quickly

1. Look at $a_n$. If it has logs, factorials, or weird mixes, Limit comparison is often easiest.
2. If $f(x)=a_x$ is pleasant to integrate and clearly decreasing, try the Integral Test.
3. For simple rational or root expressions, matching the highest power on top & bottom gives a fast $p$-series comparison.

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8. Key take‑aways

• Integral Test turns sums into areas; use when you can integrate $f(x)$.
• Direct Comparison needs an inequality; super helpful if you can show “my terms are smaller than a convergent friend” or bigger than a divergent one.
• Limit Comparison lets you skip the inequality and just check one limit.
• Memorize the behavior of geometric and $p$-series — they form the backbone of almost every comparison.