Try
A compression log has a correction term at positions 1, 2, 4, 8, 16, and so on. At one of these positions n, its magnitude is (3/5)^n; every other correction is zero. Does the sum of all correction magnitudes remain finite? Does a root test need a single ordinary root limit?
Write the first eight magnitudes and then take the nth root using the original position n. Zero has nonnegative nth root zero. No logarithm of zero is needed. These are mathematical correction terms, not a claim about any particular compressor.
A separate exact worked example, with n starting at 1| Position n | Magnitude | Nonnegative nth root |
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| 1 | 3/5 | 3/5 |
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| 2 | 9/25 | 3/5 |
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| 3 | 0 | 0 |
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| 4 | 81/625 | 3/5 |
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| 5, 6, 7 | 0 | 0 |
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| 8 | 6561/390625 | 3/5 |
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Predict
The roots alternate irregularly between two values. Predict their lower and upper limits before deciding convergence. An ordinary limit would have to agree along both the powers of two and the infinitely many other indices. Those subsequences have roots 3/5 and 0 respectively, so no ordinary root limit exists.
Nevertheless every root is at most 3/5. Therefore every magnitude is at most (3/5)^n, and the whole sum is bounded by the geometric sum (3/5)/(1−3/5)=3/2. The bound deliberately includes many positive geometric terms where the actual log contributes zero.
The important property is an eventual upper bound strictly below 1, not agreement of all root subsequences. Predict what happens if the nonzero magnitudes become 2^n instead. Sparseness does not make those terms tend to zero.
Experiment
On paper, place a horizontal threshold q=4/5 above the root values in the table. Extend the root pattern through position 32 without calculating tiny powers. Explain why the formula, rather than those 32 samples, proves that no later root crosses the threshold.
For the worked sequence, a stronger threshold 3/5 already works at every index. The absolute tail after position m is at most
(3/5)^(m+1)/(1−3/5).
At m=8 this equals 19683/781250, approximately 0.02519424. It is a bound on the omitted corrections, not their exact sum. The next occupied position is 16, so the bound can be quite conservative.
Now change the rule: put magnitude 3^(−k) at position n=k² and zero elsewhere. Calculate roots at occupied positions 1, 4, 9 and 16. They are 1/3, 3^(−1/2), 3^(−1/3), and 3^(−1/4), approaching 1. Yet the nonzero magnitudes themselves form a geometric series with sum 1/2. This second example shows why the original position matters and why the boundary value 1 needs another method. The practice task asks you to prove this conclusion, including the reindexing justification.
Understand
Let a_n be any real or complex sequence, starting at n=1. Set b_n=|a_n|^(1/n), the real nonnegative nth root of its magnitude. Define
s_N = sup of b_n over all n≥N, and L = inf over N of s_N.
Thus L=limsup b_n is an extended number between 0 and ∞. The s_N are infinite-tail suprema, not maxima of a finite screen. Two consequences of their definition drive the proof:
- If L is finite and q>L, some tail supremum is below q. Hence every sufficiently late b_n is below q.
- If q<L, infinitely many b_n exceed q. Otherwise an entire sufficiently late tail would be bounded by q and L would be at most q.
A geometric majorant when L is below 1
Choose L<q<1. There is N such that n≥N implies b_n<q. Raising nonnegative quantities to the positive integer power n gives |a_n|≤q^n. The finite initial terms have a finite sum of magnitudes, and the remaining nonnegative partial sums are bounded by a geometric series. Consequently Σ|a_n| converges, proving absolute convergence. This also proves convergence of the original real or complex series by the Cauchy criterion and the triangle inequality.
If L=0, choose any q strictly between 0 and 1; nothing in the argument requires L to be positive. For m≥N−1, the same proof gives the remainder bound q^(m+1)/(1−q). A numerical error certificate requires an established N, not just a decimal guess for L.
Failure of the term condition when L exceeds 1
Choose a fixed finite q with 1<q<L. If L=∞, q=2 is sufficient. Infinitely many indices satisfy b_n>q, so their magnitudes exceed q^n. The terms cannot approach zero. If partial sums S_n converged, their differences a_n=S_n−S_(n−1) would approach zero; this contradiction proves divergence.
For L=1 neither strict comparison is available. The series Σ1/n² converges while Σ1/n diverges, although both root sequences approach 1. The practice develops independent dyadic-block proofs instead of using the inconclusive test to support itself.
Apply
Try the six written tasks before revealing their solutions. Retain the original index when roots are taken, and distinguish three claims: the root sequence has no limit, the series diverges, and a particular test gives no answer. None of these statements alone implies either of the others.
For an additional changed example, put a_n=(1/2)^n at even indices and a_n=(1/5)^n at odd indices. The upper root limit is 1/2, so absolute convergence follows despite the missing ordinary root limit. Splitting into two geometric sums gives 1/3+5/24=13/24. Explain why splitting a nonnegative series this way is justified, and why a ratio between adjacent terms is a less useful diagnostic here.
The final task proves a power-series radius statement. For a fixed nonzero z, coefficient roots are multiplied by |z|; the root test then controls points strictly inside and strictly outside the radius. It does not determine boundary behavior. At z=0, use the constant term directly instead of forming the undefined extended product 0·∞. This separates a useful general theorem from the endpoint work that remains.
Continue
Return to the preceding tail-extrema lesson if “eventually” and “infinitely often” still seem interchangeable: the first controls every sufficiently late index, while the second requires a new witness beyond every finite cutoff. The root test uses both quantifiers for different conclusions.
For a proof record, retain your chosen q, the statement establishing its eventual range or infinitely many violations, and the conclusion it licenses. A finite table can help discover a candidate L; the sequence definition and tail argument establish it. Written practice and finite experiments remain evidence for review, not an automatic certification of a general theorem.
