Real Analysis 20 | Ratio and Root Test [dark version]

Q: Why is it not enough to show |a_{k+1}|/|a_k| < 1?

The ratio test needs a uniform bound by some fixed Q strictly less than 1 for all sufficiently large k. If the ratio merely stays below 1 but approaches 1, the terms may not shrink fast enough. The harmonic series is the classic warning: it can resemble a “ratio < 1” situation in informal reasoning yet still diverges because no fixed Q < 1 controls the tail.

Q: What makes the root test easier to apply than the ratio test?

The root test checks |a_k|^{1/k} Q for k n0. It avoids dividing by a_k, so it does not require a_k to be nonzero to form a quotient. Once |a_k|^{1/k} Q holds, raising to the kth power gives |a_k| Q^k, and Q^k is the geometric majorant.

Q: How does the root test work on a term with a power, like something involving (k+2) in the denominator?

When the term has a power structure, taking the kth root often cancels the exponent. In the example given, taking the kth root reduces the expression to a simpler fraction (the power “drops out”), producing a bound like 9/(10) for all k 8. With that uniform bound < 1, the geometric comparison applies and the series converges.

Q: What does the limit superior version of the root test add?

Instead of guessing Q and n0, compute L = limsup_{k } |a_k|^{1/k}. If L 1, the terms do not shrink fast enough and the series diverges. If L = 1, the test is inconclusive: the series could be divergent, absolutely convergent, or only conditionally convergent.

TL;DR

The ratio test gives absolute convergence if |a_{k+1}|/|a_k| is eventually bounded by a fixed Q with 0 < Q < 1.

Briefing Cornell Notes

Briefing

Ratio and root tests give fast, geometric-series-based ways to decide whether an infinite series a_k converges absolutely. The key payoff is practical: instead of hunting for a comparison majorant term, the tests turn the convergence question into checking whether a certain bound stays below 1.

For the ratio test, the method inspects the quotient |a_{k+1}|/|a_k|. If there exist an index n0 and a number Q with 0 < Q < 1 such that for all k n0 the inequality |a_{k+1}|/|a_k| Q holds (and the denominator is nonzero, so the quotient is well-defined), then the series a_k is absolutely convergent. The proof repeatedly applies the inequality to bound |a_k| by a constant times Q^{k-n0}, reducing the problem to the convergence of a geometric series. In short: once successive terms shrink by at least a factor Q < 1, the whole tail behaves like a geometric series.

A worked example uses a_k = 1/k!. Computing the ratio gives |a_{k+1}|/|a_k| = (1/(k+1)!)/(1/k!) = 1/(k+1), which is always 1/2 for k 1. Choosing Q = 1/2 (and some suitable n0) satisfies the ratio-test condition, so 1/k! converges.

The transcript also warns against a common mistake: showing |a_{k+1}|/|a_k| < 1 is not enough. The harmonic series provides the cautionary contrast—its analogous ratio-style bound can look similar but still leads to divergence, because the ratio must be bounded by some fixed Q strictly less than 1.

The root test shifts attention to the kth root of the term size: if there exist n0 and 0 < Q < 1 such that |a_k|^{1/k} Q for all k n0, then a_k is absolutely convergent. Unlike the ratio test, the root test does not require a_{k} to be nonzero for forming the quotient, making it more forgiving. The reasoning is direct: raising both sides to the kth power yields |a_k| Q^k, and Q^k is the geometric majorant.

An example illustrates the mechanics: for a_k = 9/(2+k)^{?} (as presented, the kth root cancels the power and leaves a bound like 9/(10)), the kth root test produces a uniform bound < 1 for sufficiently large k (k 8), implying convergence.

Finally, the tests can be sharpened using a limit superior. For the root test, compute limsup_{k } |a_k|^{1/k}. If this value is strictly less than 1, absolute convergence follows; if it is strictly greater than 1, divergence follows. When the limsup equals 1, the method becomes inconclusive: the series may be divergent, absolutely convergent, or merely convergent without absolute convergence. That boundary case is where other tools are needed.

Cornell Notes

Ratio and root tests decide absolute convergence of a_k by comparing term sizes to a geometric series. The ratio test checks whether |a_{k+1}|/|a_k| stays below some fixed Q with 0 < Q < 1 for all k beyond n0; then |a_k| is forced to shrink at least like Q^{k}, guaranteeing absolute convergence. The root test instead checks whether |a_k|^{1/k} Q for some Q < 1 beyond n0; it is often simpler and does not require forming a quotient, so it avoids nonzero-denominator issues. A limit superior version streamlines the root test: if limsup |a_k|^{1/k} < 1, the series converges absolutely; if it is > 1, it diverges. If the limsup equals 1, the test cannot conclude.

How does the ratio test turn a convergence problem into a geometric-series comparison?

It assumes there are n0 and Q with 0 < Q < 1 such that |a_{k+1}|/|a_k| Q for all k n0 (with |a_k| nonzero so the quotient makes sense). Multiplying the inequality repeatedly yields |a_k| bounded by a constant times Q^{k-n0}. Since Q^{k-n0} is geometric and convergent when Q < 1, the comparison majorant forces a_k to be absolutely convergent.

Why is it not enough to show |a_{k+1}|/|a_k| < 1?

The ratio test needs a uniform bound by some fixed Q strictly less than 1 for all sufficiently large k. If the ratio merely stays below 1 but approaches 1, the terms may not shrink fast enough. The harmonic series is the classic warning: it can resemble a “ratio < 1” situation in informal reasoning yet still diverges because no fixed Q < 1 controls the tail.

What makes the root test easier to apply than the ratio test?

The root test checks |a_k|^{1/k} Q for k n0. It avoids dividing by a_k, so it does not require a_k to be nonzero to form a quotient. Once |a_k|^{1/k} Q holds, raising to the kth power gives |a_k| Q^k, and Q^k is the geometric majorant.

How does the root test work on a term with a power, like something involving (k+2) in the denominator?

When the term has a power structure, taking the kth root often cancels the exponent. In the example given, taking the kth root reduces the expression to a simpler fraction (the power “drops out”), producing a bound like 9/(10) for all k 8. With that uniform bound < 1, the geometric comparison applies and the series converges.

What does the limit superior version of the root test add?

Instead of guessing Q and n0, compute L = limsup_{k } |a_k|^{1/k}. If L < 1, then the root-test inequality holds eventually for some Q < 1, so a_k converges absolutely. If L > 1, the terms do not shrink fast enough and the series diverges. If L = 1, the test is inconclusive: the series could be divergent, absolutely convergent, or only conditionally convergent.

Review Questions

For the ratio test, what exact inequality involving |a_{k+1}|/|a_k| must hold, and why must Q be strictly less than 1?
When does the root test become inconclusive, and what are the three possible outcomes in that case?
How does computing limsup_{k } |a_k|^{1/k} determine absolute convergence or divergence?

Key Points

1
The ratio test gives absolute convergence if |a_{k+1}|/|a_k| is eventually bounded by a fixed Q with 0 < Q < 1.
2
The ratio test requires the quotient to be well-defined, so the denominator |a_k| must not be zero for the relevant k.
3
A uniform bound by some Q < 1 is essential; merely having the ratio less than 1 is not enough.
4
The root test checks |a_k|^{1/k} against a fixed Q < 1 and avoids quotient/nonzero issues.
5
The root test can be automated using limsup |a_k|^{1/k}: < 1 implies absolute convergence, > 1 implies divergence.
6
If limsup |a_k|^{1/k} equals 1, the root test cannot decide; other criteria are needed.

Highlights

Absolute convergence follows once successive terms shrink by at least a fixed factor Q < 1 (ratio test) or once kth roots stay below a fixed Q < 1 (root test).

For a_k = 1/k!, the ratio |a_{k+1}|/|a_k| = 1/(k+1) quickly yields a bound like 1/2, proving convergence.

The transcript stresses a key trap: showing a ratio is merely < 1 does not guarantee convergence—Q must be uniformly bounded away from 1.

The limit superior form removes guesswork: limsup |a_k|^{1/k} < 1 forces absolute convergence, while > 1 forces divergence.

When limsup |a_k|^{1/k} = 1, the method stops short—divergence, absolute convergence, and conditional convergence all remain possible.

Topics

Ratio Test
Root Test
Absolute Convergence
Geometric Majorant
Limit Superior