\n
Problema \n<\/div>\n \n Solu\u00e7\u00e3o <\/p>\n Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n \n","protected":false},"excerpt":{"rendered":" Problema (Indicado a partir do 9\u00ba ano do E. F.) Mostre que, para todo n\u00famero natural [tex]n[\/tex] n\u00e3o nulo, vale a desigualdade [tex]\\sqrt{\\dfrac{n-1}{n}}\\lt 1[\/tex]. Utilize tal fato para determinar qual dos n\u00fameros \u00e9 o maior: [tex]200[\/tex] ou [tex]1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\dots+\\dfrac{1}{\\sqrt{10000}}[\/tex]? Solu\u00e7\u00e3o Primeiramente vamos provar que a desigualdade [tex]\\qquad \\qquad \\sqrt{\\dfrac{n-1}{n}}\\lt 1[\/tex] \u00e9 v\u00e1lida para todo n\u00famero natural … <\/p>\n
\n (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n
\nMostre que, para todo n\u00famero natural [tex]n[\/tex] n\u00e3o nulo, vale a desigualdade [tex]\\sqrt{\\dfrac{n-1}{n}}\\lt 1[\/tex].
\nUtilize tal fato para determinar qual dos n\u00fameros \u00e9 o maior: [tex]200[\/tex] ou [tex]1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\dots+\\dfrac{1}{\\sqrt{10000}}[\/tex]?<\/p>\n
\nPrimeiramente vamos provar que a desigualdade
\n[tex]\\qquad \\qquad \\sqrt{\\dfrac{n-1}{n}}\\lt 1[\/tex]
\n\u00e9 v\u00e1lida para todo n\u00famero natural [tex]n[\/tex] n\u00e3o nulo.
\nPara isso, observe que, para todo n\u00famero natural [tex]n[\/tex] n\u00e3o nulo, \u00e9 sempre verdade que [tex]n\\lt n+1[\/tex] e, al\u00e9m disso,
\n[tex]\\qquad \\qquad \\quad \\quad n \\lt n+1 \\ \\ \\Leftrightarrow \\ \\ n-1 \\lt n \\ \\ \\Leftrightarrow \\ \\ \\dfrac{n-1}{n} \\lt 1 \\ \\ \\Leftrightarrow \\ \\ \\sqrt{ \\dfrac{n-1}{n}}\\lt 1[\/tex].
\nPortanto, \u00e9 verdade que para todo n\u00famero natural [tex]n[\/tex] n\u00e3o nulo, [tex]\\boxed{\\sqrt{\\dfrac{n-1}{n}}\\lt 1}[\/tex].
\nAgora iremos transformar esta desigualdade numa outra mais \u00fatil a nossos prop\u00f3sitos:
\n[tex]\\qquad \\begin{align*}\\boxed{\\sqrt{\\dfrac{n-1}{n}}\\lt 1} &\\Leftrightarrow 1+\\sqrt{\\dfrac{n-1}{n}}\\lt 2 \\Leftrightarrow \\dfrac{\\sqrt{n}+\\sqrt{n-1}}{\\sqrt{n}}\\lt 2 \\Leftrightarrow\\\\
\n& \\Leftrightarrow\\dfrac{\\left(\\sqrt{n}+\\sqrt{n-1}\\right) \\left(\\sqrt{n}-\\sqrt{n-1}\\right)}{\\sqrt{n}} \\lt 2(\\sqrt{n}-\\sqrt{n-1}) \\Leftrightarrow \\\\
\n& \\Leftrightarrow\\boxed{ \\dfrac{1}{\\sqrt{n}} \\lt 2(\\sqrt{n}-\\sqrt{n-1})}.
\n\\end{align*}[\/tex]
\nPortanto, [tex]\\boxed{\\sqrt{\\dfrac{n-1}{n}}\\lt 1} \\Leftrightarrow\\boxed{ \\dfrac{1}{\\sqrt{n}} \\lt 2(\\sqrt{n}-\\sqrt{n-1})}.[\/tex]
\nUsando esta \u00faltima desigualdade v\u00e1rias vezes obtemos
\n[tex]\\qquad \\begin{align*}&1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\dots+\\dfrac{1}{\\sqrt{10000}}=\\dfrac{1}{\\sqrt{1}}+\\textcolor{red}{\\dfrac{1}{\\sqrt{2}}}+\\textcolor{#52D017}{\\dfrac{1}{\\sqrt{3}}}+\\dots+\\textcolor{blue}{\\dfrac{1}{\\sqrt{10000}}}\\\\
\n& \\textcolor{red}{\\boxed{\\lt}} 2(\\sqrt{1}-\\sqrt{0})+\\textcolor{red}{2(\\sqrt{2}-\\sqrt{1})}+\\textcolor{#52D017}{2(\\sqrt{3}-\\sqrt{2})}+\\cdots+\\textcolor{blue}{2(\\sqrt{10000}-\\sqrt{9999})}\\\\
\n&=(2\\sqrt{1}-2\\sqrt{0})+(2\\sqrt{2}-2\\sqrt{1})+(2\\sqrt{3}-2\\sqrt{2})+\\cdots+(2\\sqrt{10000}-2\\sqrt{9999})\\\\
\n&=-2\\sqrt{0}+2\\sqrt{10000}=0+200=200\\end{align*}\\\\
\n\\,[\/tex]
\ne, assim, [tex]1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\dots+\\dfrac{1}{\\sqrt{10000}}\\textcolor{red}{\\lt} 200.[\/tex]
\nPortanto, o maior dos dois n\u00fameros \u00e9 o [tex] \\, \\fcolorbox{black}{#eee0e5}{$200$} \\, [\/tex].<\/p>\n
\n