33. Comparison and limit comparison

The integral test needs an antiderivative — not every series' terms come from an integrable function, or the antiderivative might simply be too messy to find. This lesson trades "compute an integral" for something often easier: compare the series to another series whose behavior is already known — most often a p-series or a geometric series, both fully settled by §7.3 and §7.6.

The direct comparison test

Suppose 0\le a_n\le b_n for all n (past some starting point). - If \sum b_n converges, then \sum a_n converges. - If \sum a_n diverges, then \sum b_n diverges.

Why this works: the partial sums of a_n are trapped below the partial sums of b_n (term by term, every a_n is no bigger than the matching b_n). If b_n's partial sums are bounded (because \sum b_n converges), a_n's partial sums — being smaller and, since a_n\ge0, monotone increasing — are bounded too, so they converge by the Monotone Convergence Theorem (§7.1). The divergence direction is just the same fact read backwards: if the smaller series still blows up, the larger one has no chance of staying finite.

The test only ever runs in the direction that makes sense. Knowing \sum b_n diverges tells you nothing about the smaller \sum a_n — a tinier divergent series could easily still converge (imagine a_n shrinking much faster than b_n even while both are positive). Likewise, knowing \sum a_n converges tells you nothing about the larger \sum b_n. Only the two combinations listed above are valid.

Finding the right comparison is the actual skill. For \sum\frac1{n^2+n}, notice n^2+n>n^2, so \frac1{n^2+n}<\frac1{n^2} — and \sum\frac1{n^2} converges (p=2), so the direct comparison test concludes \sum\frac1{n^2+n} converges too.

The limit comparison test

Direct comparison demands an exact inequality, which isn't always easy to massage into place. The limit comparison test relaxes this to an asymptotic comparison instead:

Suppose a_n,b_n>0 and \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=c for some finite c>0. Then \sum a_n and \sum b_n either both converge or both diverge.

Why this works: a finite, positive limit c means that for large n, a_n\approx c\cdot b_n — the two series' terms are proportional in the long run, so multiplying b_n by the constant c (which never affects convergence — a nonzero constant multiple of a series converges exactly when the original does) sandwiches a_n between \frac c2b_n and 2cb_n for large enough n, and direct comparison closes the argument from there.

Choosing b_n: keep only the dominant terms. For \dfrac{3n+1}{n^3-2}, the numerator is dominated by 3n and the denominator by n^3 for large n, suggesting b_n=\dfrac{3n}{n^3}=\dfrac3{n^2} — or, since a constant multiple doesn't matter, simply b_n=\dfrac1{n^2}. This "keep only the fastest-growing piece top and bottom" instinct is the same one that powered §1.6's limits at infinity and §3.3's L'Hopital's rule — it resurfaces here as a way to choose a comparison, not just compute a limit.

Doing it in Python

Direct comparison, confirming the term-by-term inequality that licenses the conclusion:

n_values = range(1, 6)
print(f"{'n':>3} {'1/(n^2+n)':>12} {'1/n^2':>10}")
for n in n_values:
    a_n = 1 / (n**2 + n)
    b_n = 1 / n**2
    print(f"{n:>3} {a_n:>12.6f} {b_n:>10.6f}")
print("\na_n < b_n for every n, and sum(1/n^2) converges (p=2)")
print("=> sum(1/(n^2+n)) converges too, by direct comparison")

Limit comparison, choosing b_n by keeping only the dominant terms:

import sympy as sp

n = sp.Symbol('n', positive=True)
a_n = (3*n + 1) / (n**3 - 2)
b_n = 1 / n**2

ratio_limit = sp.limit(a_n / b_n, n, sp.oo)
print(f"a_n = {a_n}")
print(f"b_n = {b_n}   (kept only the dominant terms: 3n over n^3)")
print(f"lim (a_n / b_n) = {ratio_limit}")
print(f"finite and positive => same convergence behavior as sum(1/n^2), which converges")

A series that's awkward for direct comparison (the inequality doesn't fall out cleanly) but easy for limit comparison:

import sympy as sp

n = sp.Symbol('n', positive=True)
a_n = 1 / sp.sqrt(n**3 + 2*n)
b_n = n**sp.Rational(-3, 2)   # 1/n^(3/2), a convergent p-series (p = 3/2 > 1)

ratio_limit = sp.limit(a_n / b_n, n, sp.oo)
print(f"lim (a_n / b_n) = {ratio_limit}")
print(f"p = 3/2 > 1, so sum(b_n) converges => sum(a_n) converges too")

Worked example

Determine whether \displaystyle\sum_{n=1}^\infty\dfrac{3n+1}{n^3-2} converges, using limit comparison.

Keep the dominant terms: numerator \sim3n, denominator \sim n^3, so compare to b_n=\dfrac1{n^2} (a convergent p-series, p=2).

\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(3n+1)/(n^3-2)}{1/n^2}=\lim_{n\to\infty}\frac{n^2(3n+1)}{n^3-2}=\lim_{n\to\infty}\frac{3n^3+n^2}{n^3-2}

Divide by n^3 (§7.0's technique again):

=\lim_{n\to\infty}\frac{3+\frac1n}{1-\frac2{n^3}}=\frac{3+0}{1-0}=3

A finite, positive limit. Since \sum\frac1{n^2} converges (p=2>1):

\boxed{\sum_{n=1}^\infty\frac{3n+1}{n^3-2}\text{ converges, by limit comparison}}

Sanity check. For large n, \frac{3n+1}{n^3-2}\approx\frac{3n}{n^3}=\frac3{n^2} — visibly a constant multiple of a convergent p-series' terms, which is exactly what the limit-comparison ratio of 3 confirmed algebraically. Direct comparison would have needed to verify \frac{3n+1}{n^3-2}\le\frac{K}{n^2} for some constant K and all large n — true, but messier to pin down explicitly than simply taking a limit. ✓

Your turn

1. Use direct comparison to determine whether \displaystyle\sum_{n=1}^\infty\frac1{n^2+\sqrt n} converges (compare to a p-series).

2. Use limit comparison to determine whether \displaystyle\sum_{n=1}^\infty\frac{5n^2-1}{2n^4+n} converges.

3. True or false: if 0\le a_n\le b_n and \sum a_n converges, the comparison test guarantees \sum b_n converges too.

Solutions

1. n^2+\sqrt n>n^2, so \frac1{n^2+\sqrt n}<\frac1{n^2}. Since \sum\frac1{n^2} converges (p=2), by direct comparison \boxed{\sum\frac1{n^2+\sqrt n}\text{ converges}}.

2. Dominant terms: numerator \sim5n^2, denominator \sim2n^4, so compare to b_n=\frac1{n^2}:

\lim_{n\to\infty}\frac{(5n^2-1)/(2n^4+n)}{1/n^2}=\lim_{n\to\infty}\frac{5n^4-n^2}{2n^4+n}=\frac52

(dividing by n^4, §7.0's technique). Finite and positive; \sum\frac1{n^2} converges (p=2), so \boxed{\sum\frac{5n^2-1}{2n^4+n}\text{ converges}}, by limit comparison.

3. False. This runs the test backwards — knowing the smaller series converges says nothing about the larger one. A simple counterexample: a_n=\frac1{n^2} (converges) and b_n=\frac1n (the harmonic series, diverges), with a_n\le b_n satisfied for every n\ge1. The valid direction is exactly as stated in the concept section: a convergent larger series drags a smaller one down to convergence with it; a divergent smaller series drags a larger one up to divergence with it. The other two combinations prove nothing.

Check yourself in code

For a_n=\dfrac{3n+1}{n^3-2} and b_n=\dfrac1{n^2}, compute \displaystyle\lim_{n\to\infty}\dfrac{a_n}{b_n} with SymPy.

Print exactly this:

limit of a_n/b_n = 3

import sympy as sp

n = sp.Symbol('n', positive=True)
a_n = (3*n + 1) / (n**3 - 2)
b_n = 1 / n**2

ratio_limit = sp.limit(a_n / b_n, n, sp.oo)
print("limit of a_n/b_n = ...")

import sympy as sp

n = sp.Symbol('n', positive=True)
a_n = (3*n + 1) / (n**3 - 2)
b_n = 1 / n**2

ratio_limit = sp.limit(a_n / b_n, n, sp.oo)
print(f"limit of a_n/b_n = {ratio_limit}")

Direct comparison sandwiches a series between 0 and a known series term-by-term — smaller-than-convergent converges, larger-than-divergent diverges, and no other combination is valid. Limit comparison relaxes the exact inequality to an asymptotic one: if \frac{a_n}{b_n} settles on a finite, positive constant, the two series share a fate, which makes "keep only the dominant terms" — the same instinct §1.6 and §3.3 already built — the practical way to choose a comparison series, almost always a p-series or a geometric series from §7.3 and §7.6.

Next: what happens when a series' terms aren't all positive — alternating series, and the important distinction between a series converging outright and converging only because of cancellation.