Harmonic Series — Definition, Formula & Divergence

Harmonic Series

The series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... . Note: The harmonic series diverges. Its sequence of partial sums is unbounded.

See also

Harmonic sequence

Key Formula

\sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots = \infty

Where:

$n$ = A positive integer indexing each term of the series
$\frac{1}{n}$ = The nth term, which is the reciprocal of n

Example

Problem: Show that the harmonic series diverges by grouping its terms (the classic proof by Nicole Oresme).

Step 1: Write out the harmonic series and group terms in successive powers of 2.

1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots

Step 2: In each group, replace every term with the smallest term in that group. The group from 1/3 to 1/4 has 2 terms, each at least 1/4, so the group sums to at least 2 × 1/4 = 1/2.

\frac{1}{3} + \frac{1}{4} \geq \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Step 3: The next group has 4 terms (1/5 through 1/8), each at least 1/8, so it sums to at least 4 × 1/8 = 1/2.

\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \geq 4 \cdot \frac{1}{8} = \frac{1}{2}

Step 4: Every successive group of 2^k terms likewise contributes at least 1/2. Since there are infinitely many such groups, the total sum exceeds any finite number.

1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty

Answer: The harmonic series diverges because its partial sums can be bounded below by a sum that grows without limit.

Another Example

Problem: Compute the first few partial sums of the harmonic series and observe their slow growth.

Step 1: Compute S_1 through S_4.

S_1 = 1, \quad S_2 = 1 + \frac{1}{2} = 1.5, \quad S_3 = 1.5 + \frac{1}{3} \approx 1.833, \quad S_4 = 1.833 + 0.25 \approx 2.083

Step 2: Jump ahead: after 10 terms, S_10 ≈ 2.929. After 100 terms, S_100 ≈ 5.187. After 1,000,000 terms, S_{1000000} ≈ 14.39.

S_{10} \approx 2.93, \quad S_{100} \approx 5.19, \quad S_{1{,}000{,}000} \approx 14.39

Step 3: The partial sums grow roughly like the natural logarithm: S_n ≈ ln(n) + γ, where γ ≈ 0.5772 is the Euler–Mascheroni constant. Since ln(n) → ∞, the series diverges — but extremely slowly.

S_n \approx \ln n + 0.5772

Answer: Even after a million terms, the partial sum is only about 14.4. The harmonic series diverges, but its growth is logarithmically slow.

Frequently Asked Questions

Why does the harmonic series diverge even though its terms go to zero?

A term approaching zero is necessary for convergence but not sufficient. The terms 1/n shrink too slowly — each group of terms (when grouped by powers of 2) contributes at least 1/2, and infinitely many such groups make the total unbounded. Compare this to the series ∑1/n², whose terms shrink fast enough that the series converges to π²/6.

What is the sum of the first n terms of the harmonic series?

There is no simple closed-form expression. The nth partial sum, called the nth harmonic number H_n, is well-approximated by H_n ≈ ln(n) + γ, where γ ≈ 0.5772 is the Euler–Mascheroni constant. This approximation becomes increasingly accurate as n grows.

Harmonic Series (∑ 1/n) vs. p-Series (∑ 1/nᵖ)

The harmonic series is the specific p-series with p = 1. A p-series converges when p > 1 and diverges when p ≤ 1. So ∑1/n² (p = 2) converges to π²/6, while the harmonic series (p = 1) diverges. The harmonic series sits exactly at the boundary between convergence and divergence for p-series.

Why It Matters

The harmonic series is one of the most important examples in the study of infinite series because it shows that terms going to zero does not guarantee convergence. It appears naturally in many contexts: the expected number of coupons you need to collect to complete a set (the coupon collector problem), the overhang distance achievable by stacking blocks, and the analysis of algorithms. The related harmonic numbers H_n also arise frequently in combinatorics and number theory.

Common Mistakes

Mistake: Assuming the harmonic series converges because its terms approach zero.

Correction: Terms approaching zero is a necessary condition for convergence, not a sufficient one. You must apply a convergence test (such as the integral test or the grouping argument) to determine behavior. The harmonic series is the classic counterexample.

Mistake: Confusing the harmonic series with the alternating harmonic series.

Correction: The alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + ⋯ converges (to ln 2), while the standard harmonic series diverges. The alternating signs cause enough cancellation for convergence.

Related Terms

Series — General concept of summing infinitely many terms
Divergent Series — Category the harmonic series belongs to
Sequence of Partial Sums — Tool for analyzing series convergence
Unbounded Set of Numbers — Harmonic partial sums are unbounded
Harmonic Sequence — The sequence of terms 1/n used in the series
p-Series — Generalization where harmonic series is p = 1
Convergent Series — Contrast — what the harmonic series is not
Integral Test — Standard test proving harmonic divergence