3. Să se demonstreze că\( \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} >\frac{13}{24}, \forall n\in \mathbf N^*,n\geq 2\) Soluţie. \begin{align} &\cssId{Step1}{I. Etapa\; de\; verificare:}\\ &\cssId{Step2}{Pentru \;n=2\;:\frac{1}{3} +\frac{1}{4} =\frac{^{2)}7}{12}=\frac{14}{24} >\frac{13}{24} "A" }\\ &\cssId{Step3}{II. Etapa\; de\; demonstraţie:\; Presupunem\; adevărată}\\ &\cssId{Step4}{P(k): \frac{1}{k+1}+\frac{1}{k+2}+...+\frac{1}{2k} >\frac{13}{24}, \forall k\in \mathbf N^*,k\geq 2}\\ &\cssId{Step5}{şi\; demonstrăm\; P(k+1):\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2(k+1)} >\frac{13}{24}, \forall k\in \mathbf N^*,k\geq 2}\\ &\cssId{Step6}{\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2(k+1)} >\frac{13}{24}\vert+\frac{1}{k+1}}\\ &\cssId{Step7}{\underbrace{\left(\frac{1}{k+1}+ \frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}\right)}_{P(k)} +\frac{1}{2k+1}+\frac{1}{2k+2} >\frac{13}{24}+\frac{1}{k+1}}\\ &\cssId{Step8}{\frac{13}{24}+ \frac{1}{2k+1}+\frac{1}{2k+2} >\frac{13}{24}+\frac{1}{k+1}}\\ &\cssId{Step9}{\frac{1}{2k+1}>\frac{1}{k+1}-\frac{1}{2k+2}\Leftrightarrow \frac{1}{2k+1}>\frac{1}{2(k+1)}\; deoarece \;2k+2>2k+1\Leftrightarrow P(k+1)"A"}\\ &\cssId{Step10}{Din\; I.\; şi\; II.\;\Rightarrow P(n)"A", \forall n \in \mathbf N^*}\\ \end{align}