Comparison and Absolute Convergence

Series with terms of both signs are harder: cancellation can help or hide what is going on. The Cauchy criterion gives a clean way to compare a series with a simpler one, and it proves the fact you used in Calculus 2 that absolutely convergent series converge.

Idea clave

Comparison test. If $|a_k| \le b_k$ for all $k$ and $\sum b_k$ converges, then $\sum a_k$ converges (absolutely). If $0 \le b_k \le a_k$ and $\sum b_k$ diverges, then $\sum a_k$ diverges. Absolute convergence: if $\sum |a_k|$ converges, then $\sum a_k$ converges. A series that converges but not absolutely converges conditionally.

Por qué funciona

Let $\varepsilon > 0$. Since $\sum b_k$ converges, by the Cauchy criterion there is an $N$ with $b_{m+1} + \cdots + b_n < \varepsilon$ for $n > m \ge N$. By the triangle inequality, $$|a_{m+1} + \cdots + a_n| \le |a_{m+1}| + \cdots + |a_n| \le b_{m+1} + \cdots + b_n < \varepsilon.$$ So $\sum a_k$ satisfies the Cauchy criterion and converges. $\blacksquare$ Absolute convergence is the special case $b_k = |a_k|$. Notice what the proof needs: a convergent series above the absolute values. Being below a divergent series, or above a convergent one, gives nothing to work with.

In practice, finding the comparison series is the whole skill. Look at the terms for large $k$ and keep only the dominant parts: $\frac{k + 2}{k^{3} + 4}$ behaves like $\frac{k}{k^{3}} = \frac{1}{k^{2}}$, so compare with $\frac{C}{k^{2}}$. You do not need the best constant $C$, only one that works for all $k$ (or for all $k$ past some point, since finitely many terms never affect convergence). Bounded factors such as $\sin k$, $\cos k$ or $(-1)^{k}$ can be replaced by $1$ inside absolute values, which is why $\sum \frac{\sin k}{k^{2}}$ converges absolutely with no further work.

Worked examples

Ejemplo resuelto: Comparing with a $p$-series

Show that $\sum \frac{k + 2}{k^{3} + 4}$ converges by finding the smallest $C$ with $\frac{k + 2}{k^{3} + 4} \le \frac{C}{k^{2}}$ for all $k \ge 1$.

We need $C \ge \frac{k^{2}(k + 2)}{k^{3} + 4} = 1 + \frac{2k^{2} - 4}{k^{3} + 4}$ for every $k$.
The values for $k = 1, 2, 3, 4$ are $\frac{3}{5}$, $\frac{4}{3}$, $\frac{45}{31}$, $\frac{96}{68} = \frac{24}{17}$; after $k = 3$ the fraction $\frac{2k^{2} - 4}{k^{3} + 4}$ keeps shrinking.
So $C = \frac{45}{31}$, and $\sum \frac{C}{k^{2}}$ converges.

Respuesta: $C = \frac{45}{31}$; the series converges (absolutely).

Ejemplo resuelto: Absolutely, conditionally, or not at all

Classify $\sum \frac{(-1)^{k}}{\sqrt{k}}$, $\sum \frac{\sin k}{k^{2}}$ and $\sum (-1)^{k}\frac{k}{k + 1}$.

$\sum \frac{1}{\sqrt{k}}$ diverges ($p = \frac{1}{2}$), but the alternating series test gives convergence: conditional.
$\left|\frac{\sin k}{k^{2}}\right| \le \frac{1}{k^{2}}$: absolute.
The terms do not go to $0$: divergent.

Respuesta: Conditional; absolute; divergent.

Error común

Comparing in the wrong direction. "$0 \le a_k \le \frac{1}{k}$ and $\sum \frac{1}{k}$ diverges, so $\sum a_k$ diverges" is invalid: $a_k = \frac{1}{k^{2}}$ satisfies the hypothesis and converges. Before using a comparison, say which way it points: smaller than convergent, or bigger than divergent (with nonnegative terms). Also, the comparison must use absolute values when $a_k$ can be negative.

Worked examples

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