Virtual laboratory/Mathematics/Calculus/The harmonic series — divergence

The harmonic series — divergence

An extension of the topics of sequences, series, and limits. The sum $1+\tfrac12+\tfrac13+\cdots+\tfrac1N$ grows without bound (like $\ln N$), even though the terms $1/n$ tend to zero — strikingly different from the convergent geometric series. Nicole Oresme (c. 1350) proved the divergence by a grouping method: the condition „the terms tend to zero” is necessary, but not sufficient, for convergence.

The harmonic series — divergence
Mathematics · Mathematical analysis · Harmonic series — divergence (c. 1350)
Number of terms NN10
The sum 1+12+13+14+…1+\tfrac12+\tfrac13+\tfrac14+\dots (out to infinity) is…
terms N
10
last term 1/N
0.10000
sum H(N)
2.9290
ln N + γ
2.8798
H − (ln N + γ)
0.0492
🎯 Challenge: convergent or divergent
streak: 0

I will give you a series — decide whether it is convergent (finite sum) or divergent (sum → ∞). Remember: terms → 0 is a necessary but not sufficient condition.

What just happened?

You kept adding ever smaller fractions. Each one is tiny, and yet the partial sum kept rising steadily — and nothing stops it. This is one of the most surprising results in analysis: even though the terms 1/n1/n tend to zero, their sum tends to infinity.

The harmonic series:

H(N)=1+12+13+⋯+1NH(N) = 1+\tfrac12+\tfrac13+\dots+\tfrac1N

The partial sum H(N)H(N) grows without bound, though ever more slowly — asymptotically it behaves like the natural logarithm:

H(N)≈ln⁡NH(N)\approx\ln N

More precisely H(N)≈ln⁡N+γH(N) \approx \ln N + \gamma, where γ≈0.5772\gamma \approx 0.5772 is the Euler–Mascheroni constant. Since ln⁡N→∞\ln N \to \infty as N→∞N \to \infty, the sum has no ceiling — the series is divergent.

Why is this so surprising?

The mere fact that the terms shrink to zero does not guarantee that the sum is finite. Compare with the geometric series (the pink line on the scene):

12+14+18+⋯=1\tfrac12+\tfrac14+\tfrac18+\dots = 1

Here the terms also tend to zero, but the sum is finite (convergent, equal to 1), because they shrink fast enough. In the harmonic series the terms 1/n1/n shrink too slowly for the sum to stop. So the condition “the terms tend to zero” is necessary, but NOT sufficient for the series to converge.

Oresme’s proof (c. 1350):

The divergence was proved by the French scholar (bishop) Nicole Oresme in the 14th century with a brilliant method of grouping the terms: 12≥12\tfrac12 \ge \tfrac12, 13+14>12\tfrac13+\tfrac14 > \tfrac12, 15+16+17+18>12\tfrac15+\tfrac16+\tfrac17+\tfrac18 > \tfrac12, and so on — each successive group gives more than 12\tfrac12. There are infinitely many such groups, so the sum exceeds any number set in advance: it must be infinite.

Why this matters for your exams:

The harmonic series is the textbook counterexample: it shows that an→0a_n \to 0 does NOT imply convergence of the series ∑an\sum a_n. Remember it together with the geometric series 12+14+⋯=1\tfrac12+\tfrac14+\dots=1 as a “divergent vs convergent” pair.

Remember: 1+12+13+…1+\tfrac12+\tfrac13+\dots is divergent — the sum grows like ln⁡N\ln N and tends to infinity, even though the terms tend to zero. The terms shrinking to zero is a necessary but not sufficient condition for convergence.

Illustrative model · NexTutor Virtual laboratory

Enjoyed the experiment? Share it.

It's free — send it to friends from school.

Theory & formulas

The harmonic series

∑n=1∞1n=1+12+13+14+⋯\sum_{n=1}^{\infty}\frac1n=1+\frac12+\frac13+\frac14+\cdots The terms 1n→0\tfrac1n\to 0, and yet the sum is infinite (a divergent series).

Proof by grouping (Oresme)

13+14>12\tfrac13+\tfrac14>\tfrac12, then 15+⋯+18>12\tfrac15+\cdots+\tfrac18>\tfrac12, and so on — we keep adding infinitely many „halves”.

Rate of growth

The partial sum grows like a logarithm: ∑k=1N1k≈ln⁡N\sum_{k=1}^{N}\frac1k\approx \ln N The condition an→0a_n\to 0 is necessary, but not sufficient, for convergence.

Notation: n,k,Nn,k,N indices of terms, ln⁡\ln natural logarithm.

Frequently asked questions

›

Does the harmonic series converge?

No — it is divergent; its sum tends to infinity (though very slowly, like ln N).

›

How is this possible, given that the terms tend to zero?

The condition „the terms tend to zero” is necessary, but not sufficient, for a series to converge.

›

Who proved the divergence of the harmonic series?

Nicole Oresme in the 14th century (c. 1350), by a method of grouping the terms.

Teaching? Embed this experiment on your site.

Free widget — paste it on your school site, blog or notes. It links to the full version.

Teaching a class or tutoring? Guide students in NexTutor →