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Limit Comparison and Equivalents · Tutorial 638 of 998

Close Radicals and Rationalization

Rationalize differences of close radicals, identify the resulting power-law size, and use it to decide whether the associated positive-term series converges.

Intermediate 9 min read

What You'll Learn

  • Use the conjugate to rewrite a difference of square roots exactly.
  • Identify the size of a rationalized numerator and denominator.
  • Derive the leading order of differences of radicals with matching dominant terms.
  • Compare positive radical-difference terms with power-law series.
  • Recognize when cancellation or a sign issue invalidates a comparison.

Why Close Radicals Need Care

A difference of two large, nearby quantities can be much smaller than either quantity. This is especially important with radicals: if two square roots both grow roughly like \(n\), subtracting them directly can hide the size of their difference. In a series, that hidden size may decide whether the terms are comparable to \(1/n\), \(1/n^2\), or another power.

The main tool is rationalization: multiply a difference of square roots by the sum of those roots. The resulting identity replaces subtraction by a quotient whose numerator is the difference of the radicands. That often makes the leading order visible without needing a long expansion.

Key identity: If \(X\ge 0\), \(Y\ge 0\), and \(\sqrt X+\sqrt Y>0\), then \[ \sqrt X-\sqrt Y=\frac{X-Y}{\sqrt X+\sqrt Y}. \]

Rationalization as an Exact Step

The identity follows by multiplying by the conjugate, the sum \(\sqrt X+\sqrt Y\):

$$ \sqrt X-\sqrt Y = \frac{(\sqrt X-\sqrt Y)(\sqrt X+\sqrt Y)} {\sqrt X+\sqrt Y} = \frac{X-Y}{\sqrt X+\sqrt Y}. $$

This is an exact equality whenever the denominator is nonzero. It does not approximate the difference; it rewrites it. For a sequence of terms, the practical advantage is that \(X-Y\) may be much easier to assess than the original difference of roots.

After rationalizing, examine the numerator and denominator separately. If both radicands have the same leading growth, then their roots have the same leading growth too. The denominator is then approximately twice either root, while the numerator records how much the radicands differ. If that numerator also has cancellation, simplify it exactly before estimating its size.

A Leading-Order Rule for Nearby Radicals

Here is a useful precise version of that reasoning. Suppose \(c>0\), \(d>0\), and, as \(n\) tends to infinity,

$$ X_n\sim c n^r,\qquad Y_n\sim c n^r,\qquad X_n-Y_n\sim d n^s, $$

where \(X_n\) and \(Y_n\) are nonnegative and \(r,s\) are real numbers. For positive sequences, \(u_n\sim v_n\) means \(u_n/v_n\to 1\). Then

$$ \sqrt{X_n}-\sqrt{Y_n} \sim \frac{d}{2\sqrt c}\,n^{s-r/2}. $$

To justify the rule, rationalize first. The assumptions on \(X_n\) and \(Y_n\) give

$$ \frac{\sqrt{X_n}+\sqrt{Y_n}}{n^{r/2}} = \sqrt{\frac{X_n}{n^r}}+\sqrt{\frac{Y_n}{n^r}} \longrightarrow 2\sqrt c. $$

Also, \((X_n-Y_n)/n^s\to d\). Therefore, using the exact identity,

$$ \frac{\sqrt{X_n}-\sqrt{Y_n}}{n^{s-r/2}} = \frac{(X_n-Y_n)/n^s} {(\sqrt{X_n}+\sqrt{Y_n})/n^{r/2}} \longrightarrow \frac{d}{2\sqrt c}. $$

Because the limiting constant is positive, this limit is exactly the claimed asymptotic equivalence. The rule applies when the radicands share the same leading term and their difference has the stated leading order. If their difference is not positive, the order of the roots matters when identifying positive series terms.

Worked Series Comparisons

Worked Example: Radicands Differing by a Linear Term

Determine whether the positive-term series

$$ \sum_{n=1}^{\infty} \frac{\sqrt{n^2+3n+2}-\sqrt{n^2+n+1}}{n} $$

converges. For \(n\ge 1\), both radicands are positive, and the first exceeds the second because their difference is \(2n+1>0\). Rationalizing gives

$$ a_n = \frac{1}{n} \cdot \frac{(n^2+3n+2)-(n^2+n+1)} {\sqrt{n^2+3n+2}+\sqrt{n^2+n+1}} = \frac{2n+1} {n\bigl(\sqrt{n^2+3n+2}+\sqrt{n^2+n+1}\bigr)}. $$

Compare \(a_n\) with \(1/n\) by computing the ratio:

$$ \frac{a_n}{1/n} = \frac{2n+1}{\sqrt{n^2+3n+2}+\sqrt{n^2+n+1}} = \frac{2+1/n} {\sqrt{1+3/n+2/n^2}+\sqrt{1+1/n+1/n^2}} \longrightarrow 1. $$

The terms are positive and their ratio to \(1/n\) tends to the finite positive number \(1\). By limit comparison with the harmonic series, the given series diverges. As an algebra check, multiplying the rationalized numerator and denominator by the conjugate recovers the original difference, since the radicand difference is \(2n+1\).

Worked Example: A Fixed Difference Between Radicands

Determine whether

$$ \sum_{n=1}^{\infty} \left(\sqrt{n^2+n+1}-\sqrt{n^2+n}\right) $$

converges. The radicands are nonnegative, and the first is larger. Their difference is exactly \(1\), so rationalization gives

$$ a_n = \frac{(n^2+n+1)-(n^2+n)} {\sqrt{n^2+n+1}+\sqrt{n^2+n}} = \frac{1}{\sqrt{n^2+n+1}+\sqrt{n^2+n}}. $$

Compare with \(1/n\). The ratio is

$$ \frac{a_n}{1/n} = \frac{n}{\sqrt{n^2+n+1}+\sqrt{n^2+n}} = \frac{1}{\sqrt{1+1/n+1/n^2}+\sqrt{1+1/n}} \longrightarrow \frac12. $$

Since \(a_n>0\) and the ratio has a finite positive limit, limit comparison with \(\sum 1/n\) shows that this series diverges. The constant \(1/2\) affects the size of the terms but not the convergence decision. Rationalizing back gives \(a_n(\sqrt{n^2+n+1}+\sqrt{n^2+n})=1\), confirming the exact formula.

Worked Example: A Smaller Difference That Gives Convergence

Determine whether

$$ \sum_{n=1}^{\infty} \left(\sqrt{n^2+1+\frac1n}-\sqrt{n^2+1}\right) $$

converges. For each \(n\ge1\), both radicands are positive and the first exceeds the second by \(1/n\). Thus

$$ a_n = \frac{1/n} {\sqrt{n^2+1+1/n}+\sqrt{n^2+1}}. $$

Compare with \(1/n^2\):

$$ \frac{a_n}{1/n^2} = \frac{n} {\sqrt{n^2+1+1/n}+\sqrt{n^2+1}} = \frac{1} {\sqrt{1+1/n^2+1/n^3}+\sqrt{1+1/n^2}} \longrightarrow \frac12. $$

The terms are positive, and the ratio tends to a finite positive constant. Since the \(p\)-series \(\sum 1/n^p\) converges for \(p>1\), comparison with \(\sum 1/n^2\) proves convergence. The exact identity is checked by multiplying \(a_n\) by the sum of the two roots: the product is \(1/n\), the difference of the radicands.

Using the Power-Law Comparison

For a positive-term series, a limit comparison is useful when a ratio tends to a finite positive number. Specifically, if \(a_n>0\), \(b_n>0\) for all sufficiently large \(n\), and \(a_n/b_n\to L\) with \(0<L<\infty\), then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. The power-law model is \(b_n=1/n^p\): its series converges exactly when \(p>1\), and diverges when \(p\le 1\).

In the examples, rationalization made the comparison model visible. A fixed difference between radicands of order \(n^2\) produces a root difference of order \(1/n\). A radicand difference of order \(1/n\), while the roots still have size about \(n\), produces a root difference of order \(1/n^2\). The precise constants are found by the ratio limit, but the exponent often determines the convergence decision.

Practical rule: Rationalize first, simplify the radicand difference exactly, and only then choose a power-law comparison. Check positivity before applying limit comparison.

A Common Cancellation Mistake

A tempting but incorrect calculation is to treat a difference of square roots as the square root of the difference. For example, one might write

$$ \sqrt{n^2+1}-n \ \overset{\text{wrong}}{=}\ \sqrt{(n^2+1)-n^2} =1. $$

This is false: in general, \(\sqrt X-\sqrt Y\ne\sqrt{X-Y}\). Rationalizing gives the correct expression:

$$ \sqrt{n^2+1}-n = \frac{(n^2+1)-n^2}{\sqrt{n^2+1}+n} = \frac{1}{\sqrt{n^2+1}+n}. $$

Dividing the denominator by \(n\) shows that the term is asymptotic to \(1/(2n)\), not to \(1\). Thus the corresponding series still diverges, but for a different reason: its terms have harmonic size. The incorrect calculation misses the cancellation that rationalization exposes.

When This Method Is Useful

Close radicals appear in terms whose radicands share a dominant power of \(n\), especially when leading contributions cancel upon subtraction. Rationalization converts that cancellation into a visible difference of radicands. Once the exact difference is simplified, its order and the order of the denominator reveal the likely power-law comparison.

Do not infer convergence from a term merely tending to zero. A term of order \(1/n\) tends to zero, but its series diverges; a term of order \(1/n^2\) has a convergent comparison. Also, if the radicands do not share the same leading term, or if the radicand difference itself has hidden cancellation, do not apply the leading-order rule without checking its assumptions. Use the exact conjugate identity first, then evaluate the comparison ratio.

Check Your Understanding

Use rationalization and a power-law comparison where appropriate.

  1. Rewrite \(\sqrt{n^2+4}-\sqrt{n^2+1}\) as a quotient with no difference of radicals in the numerator.
  2. For \(\sqrt{n^2+2n+2}-\sqrt{n^2+2n+1}\), what is the exact difference of the radicands?
  3. Why must the terms be positive before applying the positive-term limit comparison test?
  4. If rationalization shows \(a_n\sim 3/n^2\), what convergence conclusion follows for \(\sum a_n\)?
  5. Explain why \(\sqrt X-\sqrt Y=\sqrt{X-Y}\) is not a valid general identity.

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