[tex]a) \frac{n - 1}{n } < \frac{n}{n + 1} [/tex]
[tex] \frac{ (n - 1)(n + 1)}{n(n + 1)} < \frac{ {n}^{2} }{n(n + 1)} [/tex]
[tex] {n}^{2} - 1 < {n}^{2} \: | - {n}^{2} [/tex]
[tex] - 1 < 0 = > relatia \: e \: adevarata[/tex]
[tex]b)( \frac{1}{2} \times \frac{3}{4} \times ... \times \frac{2n + 1}{2n} )^{2} < \frac{1}{2n + 1} [/tex]
[tex] \frac{1}{2} = \frac{2 \times 0 + 1}{2 \times 1} [/tex]
[tex] \frac{3}{4} = \frac{2 \times 1 + 1}{2 \times 2} [/tex]
[tex] \frac{5}{6} = \frac{2 \times 2 + 1}{2 \times 3} [/tex]
[tex] .............................................................[/tex]
[tex] \frac{2n + 1}{2n} = \frac{2 \times n + 1}{2 \times (n + 1)} \times \frac{n + 1}{n} [/tex]
[tex]( \frac{1}{2} ) ^{2} = \frac{1}{4} \\ \frac{1}{2n + 1} = \frac{1}{2 \times 0 + 1} = \frac{1}{3} \\ n = 0 \\ [/tex]
[tex] \frac{1}{4} < \frac{1}{3} = > (\frac{1}{2} ) ^{2} < \frac{1}{2n + 1} [/tex]
[tex] (\frac{3}{4} )^{2} = \frac{9}{16} \\ \frac{1}{2n + 1} = \frac{1}{2 \times 1 + 1} = \frac{1}{5} \\ n = 1[/tex]
[tex] \frac{9}{16} < \frac{1}{5} = > (\frac{3}{4} ) ^{2} < \frac{1}{2n + 1} [/tex]
[tex]......................................................[/tex]
[tex] (\frac{2n + 1}{2n} ) ^{2} = \frac{(2n + 1) ^{2} }{4n ^{2} } = \frac{(2n + 1) ^{3} }{4 {n}^{2}(2n + 1) } [/tex]
[tex] \frac{1}{2n + 1} = \frac{4 {n}^{2} }{4 {n}^{2} (2n + 1)} [/tex]
[tex] {(2n + 1)}^{3} < 4 {n}^{2} [/tex]
Deoarece tot termenii din stânga sunt mai mici decat cei din dreapta, atunci inecuatia este adevărată.
Sper că te-am ajutat. Succes!
