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

For $n<x<n+1$, monotonicity gives $1/x<1/n<1/(x-1)$. Integrating over that unit interval proves the strict inequalities, equivalently $\ln (1+1/n)<1/n<\ln (n/(n-1))$ for $n>1$.

Let $H_N$ be the harmonic number and define $u_N=H_N-\ln N$, $v_N=H_N-\ln (N+1)$. The same integral comparison gives $u_{N+1}-u_N=1/(N+1)-\ln (1+1/N)<0$ and $v_{N+1}-v_N=1/(N+1)-\ln (1+1/(N+1))>0$. Also $v_N<u_N$ and $u_N-v_N=\ln (1+1/N)\to 0$. The monotone convergence theorem for sequences therefore gives a common finite limit, Euler's constant $\gamma$.

The integral bounds for Euler's constant start at $v_1=1-\ln 2$ and $u_1=1$. Strict monotonicity, already at the next index, makes the limit bounds strict:Positivity is explicit because $\ln 2={\int }_1^{2}dx/x<1$. This establishes the required limit without assuming its decimal value.