Zeta parcial ζ(s,N)

Σ 1/k^s.
Created by
Renato Passos, Eng. de Software
Reviewed by
Renato Passos, Eng. de Software

Last updated: Apr 18, 2026

ζ(s)
1,642936

About this calculator

The partial zeta function calculator ζ(s,N) computes the finite sum of the Riemann zeta series up to N terms. It is defined as ζ(s,N) = Σ (1/k^s) for k from 1 to N, where 's' is a complex or real parameter and 'N' is the number of summed terms. This calculation is useful in numerical analysis, number theory, and approximations of divergent series.

The calculator's formula approximates the infinite series by truncating it at N terms. This allows estimating ζ(s) for fixed s, especially when the series converges slowly or when an initial approximation is needed. For example, with s = 2, ζ(2,N) approaches π²/6 as N increases.

Use this calculator in contexts such as convergence analysis of series, numerical algorithm testing, or special function studies. Caution: For s ≤ 1, the series diverges, and the partial sum grows indefinitely with N. Additionally, results may lose precision with very large N due to floating-point limitations.

The calculator is also relevant in mathematical physics, such as quantum field theory or statistics, where ζ(s) approximations model phenomena. However, for N approaching infinity, symbolic tools or analytical methods like Euler-Maclaurin formula are recommended for higher accuracy.

Frequently asked questions

What is the difference between the partial zeta function and the full zeta function?

The full zeta function sums infinitely many terms (N → ∞), while the partial ζ(s,N) stops at a finite N. The partial version is used for numerical approximations.

Can I use non-integer values for N?

No, N must be a positive integer. The formula requires the sum to stop exactly at k = N.

Why doesn't the calculator work for s ≤ 1?

For s ≤ 1, the series diverges (does not converge), so the partial sum grows without bound as N increases.

How does result accuracy vary with N?

Larger N yields results closer to ζ(s), but very large N may lose precision due to floating-point limitations.

Does the function support complex numbers for s?

No, this calculator uses only real s. For complex s, use specialized complex analysis tools.

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