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Integral bounds for Euler's constant (HN​−ln(N+1)<γ<HN​−lnN)

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Complex analysis Gamma function Euler's constant
2026-10-20  0 By others on same topic  0 Discussions Create my own version
For the harmonic number HN​, the upper sequence uN​=HN​−lnN decreases and the lower sequence vN​=HN​−ln(N+1) increases. The inequalities ln(1+1/N)>1/(N+1) and ln(1+1/(N+1))<1/(N+1) follow by integrating the decreasing function 1/x over a unit interval. They give the two monotonicities, while uN​−vN​=ln(1+1/N)→0. The monotone convergence theorem for sequences therefore gives the common limit Euler's constant γ. Strict monotonicity makes each finite-index bound strict. At N=1, this gives 0<1−ln2<γ<1, since ln2<1.

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  • Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 6 / b / Solution

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