A synthetic accounting process charges 1 unit for its first operation, 1/2 unit for its second, 1/3 unit for its third, and so on. The charge per operation shrinks, but does the total charge eventually stop growing? After four operations the total is
Here Hₙ, the nth harmonic sum, means the sum of 1/k for integer k from 1 through n. The index n is a positive integer; all terms are positive. The charges and indices are teaching assumptions, not timing measurements. We will estimate this total without having to add an enormous list of fractions.
Operation n
New charge 1/n
Total Hₙ
1
1
1
2
1/2
3/2
3
1/3
11/6
4
1/4
25/12
Draw a width-one rectangle of height 1/k for each term. Its area is exactly that term. The question about a sum becomes a question about areas that fit above or below the decreasing curve y=1/x.
Predict
On the interval from k to k+1, the curve begins at height 1/k and decreases. Does the rectangle with height 1/k lie above or below it? It lies above, so its area bounds the curve’s area from above. On the interval from k−1 to k, the same height is the right endpoint height, and the rectangle lies below the curve. This second comparison is only used for k≥2: the curve is not defined at zero.
Before calculating, place H₄ between ln5 and 1+ln4. Here ln is the natural logarithm. One geometric description is that ln b−ln a is the area under 1/x between positive endpoints a and b. Thus the predicted bracket is approximately 1.60944 to 2.38629, which contains 25/12. The bracket is useful but not exact; a bound is not a claim that either endpoint equals the total.
Experiment
Use the explorer to compare the reciprocal curve with its first eight rectangles. Explain one upper comparison and one lower comparison by pointing to the appropriate interval endpoints. Then change n and record the total and the two bounds. The picture keeps a fixed window; the numerical calculation can use a larger n. Do not interpret the eight visible rectangles as a picture of all terms of a much longer sum.
Predict a sum, then compare proved bounds
Hn means 1 + 1/2 + ⋯ + 1/n. Predict whether it lies below ln(n), between ln(n) and 1 + ln(n), or above both. Here ln is the natural logarithm.
No comparison revealed. Predict before revealing.
Fixed worked diagram: eight terms
This diagram always uses H8 = 761/280 ≈ 2.717857, even when the selected numerical comparison above uses another n. Axes stay fixed at x = 1 through 9 and y = 0 through 1.
On each interval [k, k + 1], the decreasing curve 1/x is at most 1/k. Eight rectangles, each width one, have total area H8. The area under the curve from 1 to 9 is ln(9), so ln(9) ≤ H8.
The two modes use the same curve and axis scales. Above or below is stated in text as well as shown by geometry. Numerical comparisons are bounded at 1,000 terms, remain temporary and award no mastery.
Make a second experiment on paper: group the terms after the initial 1 into the blocks k=2; k=3,4; k=5,6,7,8; and so on. Each new block has twice as many terms, but its smallest term has half the previous block’s smallest height. Predict whether a whole block becomes arbitrarily cheap. It does not: every such block contributes at least 1/2. This will give a divergence proof without integrals or logarithms.
The playground’s finite arithmetic helps inspect examples. The next arguments apply to every allowed integer n, and do not depend on how many examples the screen displays.
Understand
For k≥1, integrate the pointwise comparison 1/x≤1/k on [k,k+1]. Summing these adjacent intervals gives
ln(n+1) ≤ Hₙ.
For k=2 through n, compare 1/k≤1/x on [k−1,k]. Those intervals run from 1 to n, and the corresponding sum is Hₙ−1. Therefore
Hₙ ≤ 1+ln n, for every n≥1.
At n=1, the second sum is empty and its inequality is equality. No integral crosses the singular point zero. The strict decrease of the curve also makes the lower bound strict and makes the upper bound strict when n>1, although the weak inequalities are often more convenient to quote.
The same reasoning directly controls a tail. For integers 1≤m<n, the sum Hₙ−Hₘ starts at k=m+1, not at k=m. Its lower comparison covers [m+1,n+1], and its upper comparison covers [m,n]. Hence
ln((n+1)/(m+1)) ≤ Hₙ−Hₘ ≤ ln(n/m).
For example, the extra charge from operations 5 through 8 is 1/5+1/6+1/7+1/8=533/840, approximately 0.63452. It lies between ln(9/5), approximately 0.58779, and ln2, approximately 0.69315. Subtracting two full-sum lower bounds would not justify this conclusion: subtracting an independently larger quantity can reverse the desired estimate. The direct intervals give the correct proof.
For a calculus-free divergence argument, block j contains the integers strictly above 2^(j−1) through 2^j. There are 2^(j−1) terms, each at least 1/2^j. Its sum is therefore at least 1/2. Each term is also at most 1/2^(j−1), so its sum is at most 1. Thus 1+r/2 ≤ H_(2^r) ≤ 1+r. Choose r as large as needed to pass any proposed ceiling; positivity keeps every later sum above that ceiling. Terms tending to zero do not force their accumulated sum to converge.
For n≥2, divide the integral bounds by the positive number ln n. The ratio Hₙ/ln n lies between 1 and 1+1/ln n, so it tends to 1. This is the meaning of Hₙ~ln n: their relative scale agrees. It does not say their difference tends to zero.
To see the distinction precisely, put δₙ=Hₙ−ln n. Its next increment is 1/(n+1)−ln((n+1)/n), which is negative: on [n,n+1], the area under 1/x is strictly larger than the right-endpoint rectangle. Also the lower tail comparison from k=2 gives δₙ≥1−ln2+ln((n+1)/n) for n≥2. The case n=1 meets the fixed lower bound too. Since ln2<1, we have δₙ≥1−ln2>0. A decreasing bounded real sequence has a limit, so δₙ tends to a finite positive number γ. This proves existence and positivity without using a remembered decimal value of the Euler–Mascheroni constant.
Apply
Suppose a budget threshold is Hₙ>2. A sufficient certificate is ln(n+1)>2, which first holds at integer n=7 because e² is approximately 7.38906. Does that make 7 the true minimum? No: H₃=11/6 is below 2 while H₄=25/12 is above 2, so the actual minimum is 4. Positivity proves that checking these neighboring values settles the minimum. A sufficient analytic certificate may be deliberately conservative.
For another change, estimate the total of the next n terms after Hₙ. The tail bounds give ln((2n+1)/(n+1))≤H_(2n)−Hₙ≤ln2. Both endpoints tend to ln2, so doubling the index adds a total approaching ln2, even though each new term tends to zero. Explain why this agrees with the block argument rather than contradicting it.
For deeper control of the additive limit, subtract δ_N from δₙ for N>n. The resulting sum has terms ln((k+1)/k)−1/(k+1), each between zero and 1/k−1/(k+1). The latter telescope to 1/n−1/N. Let N grow: 0≤δₙ−γ≤1/n. This is an exact mathematical error bound. A decimal display additionally has rounding error; software that does not bound floating-point rounding should not describe its last printed digit as certified.
Continue
Use the written practice to derive a new tail bracket, prove a doubling-block bound and locate a threshold using exact neighboring sums. When presenting an estimate, name the quantity being bounded, the range of n and the direction of the inequality. Keep relative growth, additive error, sufficient thresholds and exact minima separate. The same rectangle strategy works for other positive decreasing functions, but each new function needs its own domain and area calculation.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
Let Hₙ=Σ from k=1 to n of 1/k for positive integers n. Compute H₅ exactly and prove ln(n+1)≤Hₙ≤1+ln n for every n≥1 by specifying the interval and rectangle height for every comparison. Explain the n=1 case and why a finite plotted sample is not the general proof.
Structure your argument and preserve revisions
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For integers 1≤m<n, prove ln((n+1)/(m+1))≤Hₙ−Hₘ≤ln(n/m). Apply it to m=4,n=7, compute the tail exactly, and explain why subtracting the two separate full-sum lower bounds is not a valid way to obtain a tail lower bound.
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Without integrals or logarithms, group the terms of H_(2^r) into blocks 2^(j−1)<k≤2^j, j=1,...,r, after the initial term 1. Prove 1+r/2≤H_(2^r)≤1+r. Deduce divergence of Hₙ for all increasing n, not only along powers of two, and give a sufficient power-of-two index for Hₙ>6.
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Find the actual least positive integer n for which Hₙ>3, using exact arithmetic near the crossing. Separately find the least integer guaranteed by the lower estimate ln(n+1)>3, and the largest integer ruled out by 1+ln n≤3. You may use e³≈20.08554 and e²≈7.38906; explain the gap between a certificate and an exact minimum.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Prove Hₙ/ln n→1 as n→∞, using n≥2. Then prove δₙ=Hₙ−ln n is decreasing and bounded below by 1−ln2>0, hence has a finite limit γ. Explain why Hₙ~ln n does not mean Hₙ−ln n→0. Do not assume a numerical value of γ.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Assume γ is the limit of δₙ=Hₙ−ln n established in the lesson. Prove 0≤δₙ−γ≤1/n without using a decimal value of γ. Use ln((k+1)/k)≤1/k and a telescoping sum. Give an interval of width at most 0.01 guaranteed to contain γ, in terms of H₁₀₀ and ln100, and distinguish it from a rounded floating-point certificate.
Structure your argument and preserve revisions
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Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
