Choosing a Power-Law Comparison Model

Why Choose a Power-Law Model?

When the terms of a positive series have several factors, it can be hard to see whether they decay quickly enough for the series to converge. A useful first step is to ask which power of \(n\) best describes their size for large \(n\).

The reason this is useful is that the behavior of a power series is already known: \(\sum 1/n^p\) converges when \(p>1\), and diverges when \(p\leq 1\). Matching a more complicated positive term to \(n^{-p}\) can therefore turn a difficult-looking series into a familiar classification problem.

Central method: Choose a power \(n^{-p}\) that captures the large-\(n\) behavior of the terms, verify the match by taking a ratio limit, and then apply the \(p\)-series criterion.

Power-Law Models and the Comparison Test

A power-law model for a positive sequence \((a_n)\) is a sequence of the form \(c n^{-p}\), where \(c>0\) and \(p\) is a real number. The constant \(c\) does not affect whether the corresponding series converges: multiplying all terms by a fixed positive constant preserves convergence or divergence.

The exponent \(p\) matters. The \(p\)-series criterion states that \(\sum_{n=1}^{\infty}1/n^p\) converges if \(p>1\), and diverges if \(p\leq1\). In particular, when \(p\leq0\), its terms do not tend to zero; when \(0<p\leq1\), it diverges by the integral test.

To transfer this classification, a ratio limit provides a precise check. Suppose \(a_n>0\), and suppose that for some \(p\),

\lim_{n\to\infty}\frac{a_n}{n^{-p}}=L \qquad\text{where }0<L<\infty.

Then \(\sum a_n\) and \(\sum 1/n^p\) have the same convergence behavior. Here is why: since the ratio tends to \(L>0\), there is an index \(N\) such that, for every \(n\geq N\),

\frac{L}{2} < \frac{a_n}{n^{-p}} < \frac{3L}{2}.

Multiplying by \(n^{-p}>0\) gives

\frac{L}{2n^p}<a_n<\frac{3L}{2n^p} \qquad(n\geq N).

Thus, from some index onward, \(a_n\) is bounded above and below by positive constant multiples of \(1/n^p\). The direct comparison test gives the same classification as the \(p\)-series. The finite number of terms before \(N\) cannot change convergence.

What the ratio check requires: The terms must be positive, and the ratio limit must be finite and strictly positive. A limit of zero or infinity does not establish this equivalence.

Choosing the Exponent

When a term contains explicit powers of \(n\), first combine their exponents using the ordinary rules of exponents. For instance, \(n^r/n^s=n^{r-s}\), so the resulting power is \(n^{-(s-r)}\). Factors that tend to a finite positive constant do not change the power-law exponent, but the ratio limit should still be checked rather than assumed.

A common source of such factors is an expression raised to a power that depends on \(n\). The following limit will be useful:

\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n=e^c \qquad(c>0\text{ fixed}).

For \(x>0\), the logarithm satisfies \(x/(1+x)\leq\ln(1+x)\leq x\). Substituting \(x=c/n\) and multiplying by \(n\) yields

\frac{c}{1+c/n} \leq n\ln\left(1+\frac{c}{n}\right) \leq c.

Both bounds tend to \(c\), so the middle expression tends to \(c\). Exponentiating proves the stated limit. This derivation makes explicit why the factor tends to a positive constant and can be used in a power-law comparison.

Worked Example: An Exponential-Looking Factor with a Convergent Power

Classify the positive series with terms

a_n=\frac{\left(1+\frac{2}{n}\right)^n}{n^{3/2}}.

The visible power suggests the model \(n^{-3/2}\). Divide by that model:

\frac{a_n}{n^{-3/2}} = \frac{\left(1+\frac{2}{n}\right)^n}{n^{3/2}}\,n^{3/2} = \left(1+\frac{2}{n}\right)^n.

Using the limit just established with \(c=2\),

\lim_{n\to\infty}\frac{a_n}{n^{-3/2}}=e^2.

Since \(0<e^2<\infty\), the ratio criterion applies. The comparison series is \(\sum 1/n^{3/2}\), which converges because \(3/2>1\). Therefore \(\sum a_n\) converges. As a check, the ratio has the positive finite limit \(e^2\), and the model exponent lies strictly above the convergence threshold.

When a Factor Tends to One

Some factors approach \(1\) even when their exponents vary with \(n\). Such a factor may leave the power-law model unchanged. To verify this, it can help to take a logarithm, calculate its limit, and then exponentiate.

Worked Example: A Slowly Varying Exponent and a Divergent Series

Classify the series with positive terms

a_n=\frac{\left(1+\frac{1}{n}\right)^{\sqrt n}}{n^{5/6}}.

Use \(n^{-5/6}\) as the candidate model. The ratio is

\frac{a_n}{n^{-5/6}} = \left(1+\frac{1}{n}\right)^{\sqrt n}.

Let \(u_n=\sqrt n\ln(1+1/n)\), the logarithm of this ratio. The inequality \(\ln(1+x)\leq x\) for \(x>0\) gives

0\leq u_n = \sqrt n\ln\left(1+\frac{1}{n}\right) \leq \sqrt n\left(\frac{1}{n}\right) = \frac{1}{\sqrt n}.

The right-hand side tends to zero, so \(u_n\to0\). Consequently,

\lim_{n\to\infty}\frac{a_n}{n^{-5/6}} = \lim_{n\to\infty}e^{u_n} = 1.

The ratio has a finite positive limit, so the series has the same behavior as \(\sum 1/n^{5/6}\). Since \(5/6\leq1\), that \(p\)-series diverges, and therefore \(\sum a_n\) diverges. The check is complete: the ratio limit is \(1\), and the comparison exponent falls in the divergent range.

Combining Powers Correctly

The exponent may not be written in the final form \(n^{-p}\) at the start. In that case, simplify the powers first, then identify \(p\). Keep the sign of the exponent visible: a term proportional to \(n^{-11/10}\), for example, corresponds to \(p=11/10\), not \(p=-11/10\).

Worked Example: Combining Powers Before Comparing

Classify the positive series with terms

a_n=\frac{n^{2/5}}{n^{3/2}\left(1+\frac{1}{n^2}\right)}.

First combine the explicit powers:

\frac{n^{2/5}}{n^{3/2}} = n^{2/5-3/2} = n^{4/10-15/10} = n^{-11/10}.

Thus the term can be written as

a_n=\frac{n^{-11/10}}{1+\frac{1}{n^2}}.

Compare with \(n^{-11/10}\):

\frac{a_n}{n^{-11/10}} = \frac{1}{1+\frac{1}{n^2}} \longrightarrow 1.

The ratio limit is positive and finite. The corresponding \(p\)-series has \(p=11/10>1\), so it converges. The original series therefore converges as well. Checking the exponent arithmetic gives \(2/5-3/2=-11/10\), and the ratio confirms that the remaining factor does not alter the classification.

A Common Mistake: Reversing the \(p\)-Series Threshold

The \(p\)-series rule is easy to reverse when thinking informally about decay. A larger \(p\) means faster decay and convergence when \(p>1\); a smaller positive \(p\) means slower decay and divergence. For example, for \(a_n=1/n^{3/4}\), the model is exactly \(n^{-3/4}\), so the ratio is \(1\). The series diverges because \(3/4<1\).

A mistaken calculation might say, “The terms still tend to zero, so the series converges.” That conclusion is false: terms tending to zero is necessary for convergence, but it is not sufficient. The correction is to use the full \(p\)-series criterion. Here \(p=3/4\leq1\), so the series diverges.

Another error is to select a power from only one factor and ignore the rest. If there are several powers of \(n\), combine all their exponents before deciding on \(p\). Then verify the candidate with a ratio. This last check guards against a factor that does not approach a finite positive constant.

A Practical Decision Process

For a positive-term series, use the following sequence of steps. It keeps the algebra, the ratio check, and the convergence decision distinct.

Identify the candidate power.
Combine explicit powers of \(n\) and rewrite the result as \(n^{-p}\), if possible.

Form the ratio.
Divide the actual term by \(n^{-p}\), keeping the terms positive and simplifying carefully.

Check the limit.
If the ratio tends to a finite positive number, the power model is valid for classification.

Apply the \(p\)-series criterion.
The series converges for \(p>1\) and diverges for \(p\leq1\).

If the ratio tends to zero or infinity, do not conclude that the original series has the same behavior as the proposed power series. That ratio does not meet the finite-positive-limit condition used here. A different comparison argument may be possible, but it requires checking the relevant inequalities and convergence direction.

Takeaway: Choosing a plausible power is only the first step. The decisive check is that the ratio to that power tends to a finite positive limit; only then does the \(p\)-series classification transfer directly.

Check Your Understanding

For each question, identify the relevant power model and justify the conclusion using the ratio and the \(p\)-series criterion.

What condition on \(\lim_{n\to\infty}a_n/n^{-p}\) allows \(\sum a_n\) to inherit the classification of \(\sum 1/n^p\)?
For \(a_n=n^{-7/5}(1+2/n)\), what is the model power, and does the series converge?
For \(a_n=n^{-2/3}\), what does the \(p\)-series criterion imply?
Why does a ratio limit of zero fail to satisfy the equivalence condition stated in this tutorial?
When simplifying \(n^{1/4}/n^{5/4}\), what value of \(p\) should be used in the comparison model?