Harmonic Series Calculator

Sum consecutive terms of the harmonic series — 1/a + 1/(a+1) + ... + 1/(a+n−1) — and see the exact partial sum alongside its natural-log based estimate.

Quick Facts

Formula
H(a→b) = Σ 1/k for k = a to a+n−1
Equals H(a+n−1) − H(a−1); for large m, H(m) ≈ ln(m) + γ, where γ ≈ 0.5772156649 (Euler–Mascheroni constant). The series diverges as n → ∞, but only as fast as ln(n).

Your Results

Calculated
Partial sum
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Σ 1/k from term a to a+n−1
Last term added
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1 / (a+n−1), the smallest addend
Average term value
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Partial sum ÷ number of terms
Asymptotic estimate
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ln-based approximation (Euler–Mascheroni)

Ready

Set the starting term and number of terms, then press Calculate.

About the Harmonic Series

The harmonic series is the sum of reciprocals of the positive integers: 1 + 1/2 + 1/3 + 1/4 + 1/5 + .... Each individual term shrinks toward zero, but the running total never settles down — it keeps climbing forever as more terms are added. This calculator sums any consecutive block of terms, from a chosen starting integer a through n terms, and reports the exact partial sum plus a fast logarithm-based estimate.

The formula

The partial sum from term a through n terms is:

H(a → b) = 1/a + 1/(a+1) + 1/(a+2) + ... + 1/b, where b = a + n − 1

When a = 1, this is the ordinary harmonic number Hn. For any other starting point, the sum equals the difference of two harmonic numbers: H(a→b) = Hb − Ha−1. That identity is why summing "term 500 through term 600" gives the same answer whether you sum those 101 terms directly or subtract H₄₉₉ from H₆₀₀.

Why the series diverges

It is tempting to assume that because each added term 1/k gets smaller, the total must approach some finite limit — but it does not. The classic proof, attributed to the 14th-century scholar Nicole Oresme, groups terms into blocks that each sum to at least 1/2: (1/3 + 1/4) > 1/2, (1/5 + 1/6 + 1/7 + 1/8) > 1/2, and so on forever. Since you can always find another block worth at least 1/2, the running total has no ceiling. The catch is speed: reaching a partial sum of just 20 requires roughly 272 million terms, because the series grows only like the natural logarithm of n.

The Euler–Mascheroni approximation

For large m, the harmonic number Hm is very well approximated by ln(m) + γ + 1/(2m) − 1/(12m²), where γ ≈ 0.5772156649 is the Euler–Mascheroni constant — the limiting gap between the harmonic sum and the natural logarithm. This calculator computes the exact partial sum by direct addition and also reports this asymptotic estimate, so you can see how closely the two track for your chosen range.

Reading the results

  • Partial sum is the exact total of the n terms you selected, computed by direct summation.
  • Last term added shows how small the final addend (1/b) has become — a reminder that shrinking terms are not the same as a shrinking sum.
  • Average term value is the partial sum divided by n, useful for comparing ranges of different lengths.
  • Asymptotic estimate uses the ln(m) + γ formula to sanity-check the exact sum without adding up every term by hand.

Frequently Asked Questions

What is the harmonic series?
The harmonic series is the sum of reciprocals of the positive integers: 1 + 1/2 + 1/3 + 1/4 + ... + 1/n. Each finite partial sum H(n) is a normal, finite number, but as n grows without bound the total grows without bound too — the series diverges. It just does so extremely slowly, at roughly the rate of the natural logarithm of n.
Does the harmonic series converge?
No. Although each term 1/k gets smaller and smaller, the sum still grows past any bound as more terms are added — this was proven by Nicole Oresme in the 14th century using a grouping argument. It is a classic example showing that terms shrinking to zero is not enough to guarantee a series converges.
How is the harmonic sum estimated for large n?
For large n, the partial sum H(n) is closely approximated by ln(n) + γ, where γ is the Euler–Mascheroni constant, approximately 0.5772156649. Adding the correction terms 1/(2n) − 1/(12n²) makes the estimate accurate to many decimal places even for moderate n.
Can I sum a range that does not start at 1?
Yes. This calculator sums n consecutive terms starting from any positive integer a, computing 1/a + 1/(a+1) + ... + 1/(a+n−1). This equals the difference of two harmonic numbers, H(a+n−1) − H(a−1), which is useful for checking partial ranges without re-summing from the beginning.

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