Comparison Test and Square-Summable Sequences

The comparison test

For series with nonnegative terms, convergence is governed entirely by size. The direct comparison test says: suppose $0 \le a_n \le b_n$ for all large $n$. Then

$\sum b_n \text{ converges} \;\Rightarrow\; \sum a_n \text{ converges},\qquad \sum a_n \text{ diverges} \;\Rightarrow\; \sum b_n \text{ diverges}.$

In words: a series squeezed below a convergent one converges; a series sitting above a divergent one diverges. Only the tail matters — changing finitely many terms never changes whether a series converges.

A constant factor is harmless: if $0 \le a_n \le C\,b_n$ for some fixed $C > 0$ and $\sum b_n$ converges, then $\sum a_n \le C\sum b_n < \infty$ converges as well. Spotting that bounding constant is usually the whole game.

Limit comparison (the easy companion)

When a clean inequality is fiddly, use the limit comparison test. If $a_n, b_n > 0$ and

$\lim_{n\to\infty} \frac{a_n}{b_n} = L \quad\text{with}\quad 0 < L < \infty,$

then $\sum a_n$ and $\sum b_n$ share the same fate — both converge or both diverge. A finite nonzero limit means the two series have the same order of magnitude.

Two facts you keep reusing

• Term test (necessary condition): if $\sum a_n$ converges then $a_n \to 0$. So $\sum a_n^2 < \infty$ forces $a_n^2 \to 0$, hence $a_n \to 0$.
• Square-summable $\Rightarrow$ a lot: a sequence with $\sum a_n^2 < \infty$ is called square-summable ($\ell^2$). Its terms must vanish, and any expression dominated by a constant times $a_n^2$ also forms a convergent series.

How to apply (checklist)

1. Confirm nonnegativity of the terms (eventually). The comparison test needs $a_n \ge 0$.
2. Guess the dominating series $b_n$ — pick the convergent benchmark you already trust ($\sum a_n^2$, a $p$-series with $p>1$, or a geometric series).
3. Bound the tail: show $0 \le a_n \le C\,b_n$ for all large $n$, often by trapping a slowly-varying factor (like $\tfrac{1}{1+a_n}$) inside a constant interval.
4. Conclude convergence (or, with the reverse inequality, divergence).

Worked mini-example

Suppose $-1 < a_n < 0$ and $\sum a_n^2$ converges. Does $\displaystyle\sum \frac{a_n^2}{1+a_n}$ converge?

Since $\sum a_n^2$ converges, $a_n^2 \to 0$, so $a_n \to 0$. Then $1 + a_n \to 1$, so for all large $n$ we have $1 + a_n > \tfrac12$, giving $\dfrac{1}{1+a_n} < 2$. Therefore

$0 < \frac{a_n^2}{1+a_n} = a_n^2 \cdot \frac{1}{1+a_n} < 2\,a_n^2.$

The terms are nonnegative and bounded by $2\,a_n^2$, whose series $2\sum a_n^2$ converges. By the comparison test, $\displaystyle\sum \frac{a_n^2}{1+a_n}$ converges.

The same constant-bounding trick handles $\sum |a_n|^3$, $\sum \frac{a_n^2}{1-a_n}$, and similar tails: once a factor is pinned inside a fixed interval, it collapses to a harmless constant $C$ in front of the trusted $\sum a_n^2$.

Exam trick: To use the comparison test you must first get the unruly factor under control — show $a_n \to 0$, then trap $\tfrac{1}{1+a_n}$ (or whatever varies) inside a constant interval for large $n$. The 'finitely many bad terms' near the start can be ignored; convergence is decided purely by the tail. A distractor that claims divergence usually forgets this and overreacts to the first few terms.