Question

mostre por indução que
1/n+1+1/n+2+......+1/2n>13/24,para todo n>1

Answer 1

Queremos mostrar por indução que

$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\ldots+\dfrac{1}{2n}>\dfrac{13}{24}\\ \\ \\ \boxed{\displaystyle\sum\limits_{p=n+1}^{2n}~{\dfrac{1}{p}}>\dfrac{13}{24}}$

para todo $n>1$ natural.

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$\bullet\;\;$ Caso base. Para $n=2:$

$\displaystyle\sum\limits_{p=2+1}^{2\,\cdot\, 2}~{\dfrac{1}{p}}\\ \\ \\ =\sum\limits_{p=3}^{4}{\dfrac{1}{p}}\\ \\ \\ =\dfrac{1}{3}+\dfrac{1}{4}\\ \\ \\ =\dfrac{4}{12}+\dfrac{3}{12}\\ \\ \\ =\dfrac{7}{12}=\dfrac{14}{24}>\dfrac{13}{24}~~~~~~(\checkmark)$


$\bullet\;\;$ Suponhamos que a proposição vale para $n=k-1>2:$

$\displaystyle\sum\limits_{p=(k-1)+1}^{2\,(k-1)}~{\dfrac{1}{p}}>\dfrac{13}{24}\\ \\ \\ \sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}}>\dfrac{13}{24}~~~~~~\text{(hip\'{o}tese de indu\c{c}\~{a}o)}$


$\bullet\;\;$ Para $n=k,$ temos

$\displaystyle\sum\limits_{p=k+1}^{2k}~{\dfrac{1}{p}}\\ \\ \\ =\left(\sum\limits_{p=k+1}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{2k-1}+\dfrac{1}{2k}$


Somando e subtraindo $\dfrac{1}{k},$ temos

$=\displaystyle\left(\sum\limits_{p=k+1}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{2k-1}+\dfrac{1}{2k}+\dfrac{1}{k}-\dfrac{1}{k}\\ \\ \\ =\left(\dfrac{1}{k}+\sum\limits_{p=k+1}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{2k-1}+\dfrac{1}{2k}-\dfrac{1}{k}\\ \\ \\ =\left(\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{2k-1}+\dfrac{1}{2k}-\dfrac{1}{k}$


Reduzindo as frações fora dos parênteses ao mesmo denominador, temos

$=\displaystyle\left(\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{(2k)+(2k-1)-2\cdot (2k-1)}{(2k-1)\cdot (2k)}\\ \\ \\ =\left(\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{2k+2k-1-4k+2}{(2k-1)\cdot (2k)}\\ \\ \\ =\left(\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{(2k-1)\cdot (2k)}~~~~~~\mathbf{(i)}$


Como $\dfrac{1}{(2k-1)\cdot (2k)}>0\,,$ então

$\displaystyle\left(\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}} \right )+\dfrac{1}{(2k-1)\cdot (2k)}>\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}}>\dfrac{13}{24}\\ \\ \\ \sum\limits_{p=k+1}^{2k}~{\dfrac{1}{p}}>\sum\limits_{p=k}^{2k-2}~{\dfrac{1}{p}}>\dfrac{13}{24}\\ \\ \\ \boxed{\sum\limits_{p=k+1}^{2k}~{\dfrac{1}{p}}>\dfrac{13}{24}}~~~~~~(\checkmark)$

como queríamos demonstrar.