Why Radical Terms Need Bounds
A radical in a denominator can make a term look unfamiliar even when its size is controlled by a simple power of \(n\). For example, the denominator \(\sqrt{n^2+n}\) contains two powers of different degrees, but for positive integers \(n\) it lies between constant multiples of \(n\). That observation lets us compare its reciprocal with the harmonic terms \(1/n\).
The goal is not to simplify every radical into an exact expression. Instead, find a useful inequality and choose a comparison series whose behavior is known. The p-series criterion gives the benchmark: \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\) and diverges when \(p\leq 1\).
The inequality direction is essential. A smaller series can converge even when a larger one diverges, so an upper bound by a divergent series proves nothing about convergence. Likewise, a lower bound by a convergent series does not establish divergence.
Useful Inequalities for Radicals
For nonnegative \(x\) and \(y\), the square root is increasing, and therefore \(x\leq y\) implies \(\sqrt{x}\leq\sqrt{y}\). A particularly useful pair of bounds is
For the upper bound, both sides are nonnegative, so it is valid to square them. The right-hand square is \(x+y+2\sqrt{xy}\), which is at least \(x+y\). Thus \(\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}\). The lower bound follows from \(x\leq x+y\) and the fact that square root preserves order.
When the terms inside a radical have comparable size, another convenient bound is \(\sqrt{x+y}\leq\sqrt{2y}\) if \(0\leq x\leq y\). Indeed, \(x+y\leq2y\), and taking square roots preserves the inequality. These bounds often turn a radical expression into a constant times a power of \(n\).
For positive quantities, taking reciprocals reverses an inequality: if \(0<u\leq v\), then \(1/v\leq1/u\). This reversal is especially important when the radical appears in a denominator.
Worked Example: A Square-Root Denominator That Gives Divergence
Worked Example: A Square-Root Denominator That Gives Divergence
Determine whether the series with general term \(a_n=1/\sqrt{n^2+n}\) converges.
For every \(n\geq1\), we have \(n\leq n^2\). Hence
All quantities are nonnegative, so taking square roots gives
The denominator is positive. Taking reciprocals therefore reverses the inequality:
The comparison series \(\sum 1/n\) is the divergent p-series with \(p=1\), and multiplying its terms by the positive constant \(1/\sqrt{2}\) does not change divergence. Since \(a_n\) is at least those divergent comparison terms, the direct comparison test shows that \(\sum a_n\) diverges.
Check: The bound was obtained from \(n^2+n\leq2n^2\); after taking square roots and reciprocals it gives \(a_n\geq1/(\sqrt{2}n)\), the lower bound required for divergence.
Radical Terms That Are Bounded Above by a Convergent p-Series
For convergence, the useful move is often to discard a positive part of the denominator. Doing so makes the denominator smaller and the fraction larger, so the resulting expression is an upper bound on the original term. If that larger bound is already a convergent p-series term, convergence follows.
Worked Example: A Radical Added to a Cubic Denominator
Determine whether the series with general term \(a_n=n/(n^3+\sqrt{n})\) converges.
For \(n\geq1\), \(\sqrt{n}>0\), so \(n^3+\sqrt{n}\geq n^3>0\). The numerator is positive. Dividing by the larger denominator gives a smaller fraction:
The comparison series \(\sum 1/n^2\) is a p-series with \(p=2>1\), so it converges. The direct comparison test therefore proves that \(\sum n/(n^3+\sqrt{n})\) converges.
Check: The denominator inequality \(n^3+\sqrt{n}\geq n^3\) yields an upper bound, \(a_n\leq1/n^2\), and the comparison series is convergent. Both the direction and the required p-series behavior are correct.
More Than Square Roots
The same reasoning works for other roots. For example, the cube-root function preserves order on nonnegative numbers: if \(u\leq v\), then \(\sqrt[3]{u}\leq\sqrt[3]{v}\). This allows comparisons with powers just as for square roots.
Worked Example: A Cube-Root Denominator and Harmonic Divergence
Determine whether the series with general term \(a_n=1/\sqrt[3]{n^3+n}\) converges.
For \(n\geq1\), \(n\leq n^3\). Thus
Taking cube roots preserves order, giving
The denominator is positive, so taking reciprocals reverses the inequality:
The harmonic series diverges, as does any positive constant multiple of it. The terms \(a_n\) are at least that constant multiple, so the original series diverges by direct comparison.
Check: The cube-root bound follows from \(n^3+n\leq2n^3\), and reciprocation produces the required lower bound by a constant times \(1/n\). Therefore divergence is the conclusion supported by the inequality.
A Radical Bound That Proves Divergence by a Different Route
Sometimes a denominator contains both a radical and another term. Bound each part separately, then combine the bounds to obtain a convenient estimate for the whole denominator.
Worked Example: Bounding a Sum in the Denominator
Determine whether the series with general term \(a_n=1/(\sqrt{n^2+4n}+n)\) converges.
For \(n\geq1\), \(4n\leq8n^2\), so \(n^2+4n\leq9n^2\). Taking square roots gives \(\sqrt{n^2+4n}\leq3n\). It follows that
Since the denominator is positive, reciprocals reverse the inequality:
The comparison series \(\sum 1/(4n)\) is one quarter of the harmonic series and diverges. The given series therefore diverges by direct comparison.
Check: The denominator is at most \(4n\), so the term is at least \(1/(4n)\). This is a divergent lower comparison, as required.
How to Choose and Verify a Comparison
A comparison is useful only if it has both the right direction and a known series behavior. A reliable procedure is to establish positivity, simplify the radical inequality, reverse the inequality if taking reciprocals, and then identify the exponent in the p-series.
Confirm that the terms and denominators are positive on the index range being used, so inequalities and reciprocals are valid.
Use inequalities such as \(n\leq n^2\) for \(n\geq1\), then take the appropriate root, which preserves order for nonnegative quantities.
A larger positive denominator gives a smaller reciprocal. Write this step explicitly rather than relying on intuition.
For convergence, use an upper bound by a convergent p-series. For divergence, use a lower bound by a divergent p-series.
It is enough for the termwise inequality to hold for all sufficiently large \(n\). If it starts at \(n=N\), the omitted terms before \(N\) form a finite sum. Adding or removing finitely many finite terms cannot change whether an infinite series converges.
Common Pitfall: Reversing the Comparison Logic
For \(n\geq1\), \(\sqrt{n^2+n}\geq n\), so one valid calculation is
A mistaken conclusion would be: “The harmonic series diverges, so the series with terms \(1/\sqrt{n^2+n}\) diverges.” That conclusion does not follow. The inequality makes the radical terms smaller than the harmonic terms; a smaller positive series may converge even if the larger comparison series diverges.
To prove divergence, use the other bound \(n^2+n\leq2n^2\). It gives
which is a lower bound by a divergent p-series. This is the comparison that supports the divergence conclusion.
Why These Comparisons Matter
Radical expressions often conceal an ordinary power-law scale. Once a term is bounded by a constant multiple of \(1/n^p\), the p-series threshold gives a clear test: exponents above \(1\) yield a convergent comparison, while exponents at or below \(1\) yield a divergent comparison. The constant does not change that threshold.
Careful bounds also show exactly what has been proved. A lower bound by a divergent series certifies that the original terms do not decrease enough for convergence; an upper bound by a convergent series controls the total size of the original terms. The argument depends on inequalities, not merely on the impression that two expressions “look similar.”
Check Your Understanding
For each question, identify the comparison inequality and the direction it must have.
- For \(n\geq1\), show that \(1/\sqrt{n^2+3n}\geq1/(2n)\), and state what this implies about the corresponding series.
- Explain why \(n/(n^3+\sqrt{n})\leq1/n^2\) is useful for proving convergence.
- If a positive term satisfies \(a_n\leq1/n\), does that alone prove that \(\sum a_n\) diverges? Explain.
- For \(n\geq1\), find an upper bound for \(\sqrt{n^2+9n^2}\) that is a constant times \(n\).
- State the comparison direction needed to prove divergence using the harmonic series.
