What is the asymptotic comparison test

The asymptotic comparison test compares two series with positive terms through the limit of their ratio rather than through a term-by-term inequality. It removes the main difficulty of the direct comparison test, where one must guess the right inequality and prove it. Let $\sum a_n$ and $\sum b_n$ be two series with positive terms, and suppose the terms are asymptotically equivalent:

$$ a_n \sim b_n \quad \text{as } n \to +\infty $$

The two series then share the same behavior, they either both converge or both diverge. The relation $a_n \sim b_n$ means that the ratio of the terms tends to $1$:

$$ \lim_{n \to +\infty} \frac{a_n}{b_n} = 1 $$

The conclusion holds more generally whenever this limit is finite and strictly positive, since a ratio that stabilizes at a positive constant keeps the two general terms of the same order of magnitude.

Asymptotic equivalence concerns only the eventual size of the terms, so the test reads the nature of a series from a simpler model series with the same dominant behavior.

Why it works

Suppose $\frac{a_n}{b_n} \to L$ with $0 < L < +\infty$. From the definition of limit, for every small $\varepsilon > 0$ an integer $N$ exists such that for all $n \geq N$:

$$ (L - \varepsilon) b_n < a_n < (L + \varepsilon) b_n $$

Both inequalities involve only positive multiples of $b_n$. If $\sum b_n$ converges, the right inequality bounds $a_n$ above by a convergent series, so $\sum a_n$ converges by the direct comparison test. If $\sum b_n$ diverges, the left inequality bounds $a_n$ below by a divergent series, so $\sum a_n$ diverges. The two series therefore have the same nature.

A natural reference for this comparison is the generalized harmonic series, whose convergence is fully known in terms of its exponent. Another frequent model is the Mengoli series $\sum \frac{1}{n(n+1)}$, which converges to $1$ and is asymptotically equivalent to $\sum \frac{1}{n^2}$, since $\frac{1}{n^2} \sim \frac{1}{n(n+1)}$.

Example

Using the asymptotic comparison test, we study the nature of the series:

$$ \sum_{n=1}^{\infty} \frac{e^{\sin^3\left(\frac{1}{\sqrt[3]{n}}\right)} - 1}{\sqrt[3]{n}} $$

The terms are positive for every $n \geq 1$, so a comparison with a positive-term model is legitimate. As $n \to +\infty$ the argument $\frac{1}{\sqrt[3]{n}}$ tends to zero, which lets us replace each factor by its leading-order behavior. Using the remarkable limits $\sin x \sim x$ and $e^x - 1 \sim x$ for $x \to 0$, the numerator satisfies:

$$ e^{\sin^3\left(\frac{1}{\sqrt[3]{n}}\right)} - 1 \sim \sin^3\left(\frac{1}{\sqrt[3]{n}}\right) \sim \left(\frac{1}{\sqrt[3]{n}}\right)^3 = \frac{1}{n} $$

Dividing by $\sqrt[3]{n}$ gives the model term:

$$ a_n \sim \frac{1}{n \sqrt[3]{n}} = \frac{1}{n^{1 + \frac{1}{3}}} = \frac{1}{n^{\frac{4}{3}}} = b_n $$

The series $\sum b_n$ is a generalized harmonic series with exponent $\frac{4}{3} > 1$, so it converges. By the asymptotic comparison test the original series converges as well.