= 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 $u_N-v_N=\ln(1+1/N)\to0$. 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 $v_1=1-\ln2$ and $u_1=1$. Strict monotonicity, already at the next index, makes the limit bounds strict:
$$
\boxed{0<1-\ln2<\gamma<1.}
$$
Positivity is explicit because $\ln2=\int_1^2dx/x<1$. This establishes the required limit without assuming its decimal value.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2-harmonic-logarithmic-bounds.png]
{title=Integral comparison and nested bounds for Euler's constant}
{description=Original graph: the height $1/n$ lies between $1/x$ and $1/(x-1)$ on $[n,n+1]$, illustrated for $n=2$. The second panel shows the decreasing upper and increasing lower sequences; their gap tends to zero.}
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