Consider the following functions where n is a positive integer: n1/3, log⁡(n),…

Concept

Asymptotic growth compares how positive functions scale as n becomes large. Every fixed positive power of n eventually dominates a logarithm, and Stirling’s approximation gives log(n!) = Θ(n log(n)).

The stated base matters here: with base 2, 2log₂(n) = n.

Application

1. A logarithm is dominated by any positive polynomial power, so log2(n) = o(n1/3).

2. Using the base-2 identity, 2log₂(n) = n. Since n1/3 = o(n), the fractional-power term comes before this exponential-log term.

3. Stirling’s approximation gives log2(n!) = Θ(n log₂(n)). Therefore n = o(log₂(n!)).

Cross-check

• log₂(n) / n1/3 → 0.

• n1/3 / n = n−2/3 → 0.

• n / (n log₂(n)) = 1 / log₂(n) → 0.

Result

log2(n) < n1/3 < 2log₂(n) < log2(n!).