Comparison Tests

If you cannot evaluate a series, compare it to one you understand. A series with smaller positive terms cannot sum to more than a convergent one above it; a series with larger terms cannot sum to less than a divergent one below it. That single idea — squeeze your unknown series between a known one and infinity — gives the two most-used convergence tests.

0 \le a_n \le b_n: \qquad \sum b_n \text{ converges} \Rightarrow \sum a_n \text{ converges}, \qquad \sum a_n \text{ diverges} \Rightarrow \sum b_n \text{ diverges}.

The trick is finding the right yardstick. Most often it is a geometric series or a p-series — the two families whose fate you already know cold.

The Direct Comparison Test, worked line by line

Suppose 0 \le a_n \le b_n for all n, and that the larger series \sum b_n converges. We show \sum a_n converges.

Step 1 — partial sums are increasing. Every term a_n \ge 0, so adding more terms only grows the partial sum s_n = \sum_{k=1}^{n} a_k:

s_1 \le s_2 \le s_3 \le \cdots

Step 2 — bound them above, term by term. Since a_k \le b_k for each k, the sum of the small terms is at most the sum of the big ones, which is at most the whole convergent total B = \sum_{k=1}^{\infty} b_k:

s_n = \sum_{k=1}^{n} a_k \;\le\; \sum_{k=1}^{n} b_k \;\le\; \sum_{k=1}^{\infty} b_k = B.

Step 3 — invoke monotone convergence. The sequence (s_n) is increasing (Step 1) and bounded above by B (Step 2). An increasing sequence with a ceiling must converge to a limit no larger than that ceiling:

\sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} s_n \le B < \infty.

The divergence half is the contrapositive: if \sum a_n = \infty, the bigger \sum b_n cannot be finite either.

A concrete direct comparison

Test \sum_{n=1}^{\infty} \dfrac{1}{n^2 + 1}. For every n the denominator is bigger than n^2, so the term is smaller than the corresponding p-series term:

0 \le \frac{1}{n^2 + 1} \le \frac{1}{n^2}.

The yardstick \sum 1/n^2 converges (it is a p-series with p = 2 > 1), so by direct comparison the smaller series converges too.

The Limit Comparison Test, worked line by line

Direct comparison needs a clean inequality, which is fiddly when the terms only behave like a known series. The limit comparison test removes the fuss: if the terms are eventually proportional, the series share a fate. Let a_n, b_n > 0 and suppose

\lim_{n\to\infty} \frac{a_n}{b_n} = c, \qquad 0 < c < \infty.

Step 1 — turn the limit into a two-sided bound. The ratio settling on a positive c means that, past some index N, the ratio is trapped in a band around c — say between \tfrac{c}{2} and 2c:

\frac{c}{2} \;<\; \frac{a_n}{b_n} \;<\; 2c \qquad \text{for all } n \ge N.

Step 2 — multiply through by b_n > 0. This sandwiches the terms a_n between two constant multiples of b_n:

\frac{c}{2}\, b_n \;<\; a_n \;<\; 2c\, b_n \qquad \text{for all } n \ge N.

Step 3 — apply direct comparison both ways. A nonzero constant multiple does not change whether a series converges, so:

\sum b_n \text{ converges} \;\overset{a_n < 2c\,b_n}{\Longrightarrow}\; \sum a_n \text{ converges}, \qquad \sum b_n \text{ diverges} \;\overset{a_n > \frac{c}{2}\,b_n}{\Longrightarrow}\; \sum a_n \text{ diverges}.

Step 4 — conclude the shared fate. Both implications run, so the two series stand or fall together:

\sum a_n \text{ converges} \;\iff\; \sum b_n \text{ converges}.

A concrete limit comparison

Test \sum_{n=1}^{\infty} \dfrac{3n + 5}{n^3 - 2n + 4}. For large n the term behaves like 3n / n^3 = 3/n^2, so compare with b_n = 1/n^2:

\frac{a_n}{b_n} = \frac{(3n+5)/(n^3 - 2n + 4)}{1/n^2} = \frac{(3n + 5)\,n^2}{n^3 - 2n + 4} = \frac{3n^3 + 5n^2}{n^3 - 2n + 4} \;\xrightarrow[n\to\infty]{}\; 3.

The limit c = 3 is finite and positive, and \sum 1/n^2 converges — so the original series converges too, no inequality-juggling required.

Let a_n, b_n \ge 0. Then:

Direct comparison. If 0 \le a_n \le b_n for all n, then \sum b_n converges \Rightarrow \sum a_n converges, and \sum a_n diverges \Rightarrow \sum b_n diverges.
Limit comparison. If a_n, b_n > 0 and \displaystyle\lim_{n\to\infty} a_n/b_n = c with 0 < c < \infty, then \sum a_n and \sum b_n both converge or both diverge.
Edge cases. If c = 0 only the "converges" direction survives (\sum b_n conv. \Rightarrow \sum a_n conv.); if c = \infty only the "diverges" direction survives.

The art is picking b_n. The reliable instinct: keep only the fastest-growing term on top and bottom, and let the rest fall away. For a ratio of polynomials, that leaves a power of n:

a_n = \frac{3n + 5}{n^3 - 2n + 4} \;\approx\; \frac{3n}{n^3} = \frac{3}{n^2} \;\sim\; \frac{1}{n^2}.

So compare with the p-series 1/n^2; the leftover constant 3 is exactly the finite, positive c that limit comparison wants. The rule of thumb: a rational term of degree d on the bottom over degree e on top behaves like 1/n^{\,d - e}, so it converges iff d - e > 1. This is why limit comparison is the workhorse for "messy fractions of polynomials".

Watch one series bound another

The faint curve is the convergent yardstick b_n = 1/n^2; the bold curve is a_n = 1/(n^2 + k), which sits at or below it for every k \ge 0. The second chart shows their partial sums — the small series' total stays pinned beneath the convergent one, exactly as direct comparison promises.