{"id":11881,"date":"2014-05-22T00:23:13","date_gmt":"2014-05-21T23:23:13","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11881"},"modified":"2022-01-13T01:37:07","modified_gmt":"2022-01-13T01:37:07","slug":"prove-que-a-sucessao-e-limitada","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11881","title":{"rendered":"Prove que a sucess\u00e3o \u00e9 limitada"},"content":{"rendered":"<p><ul id='GTTabs_ul_11881' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11881' class='GTTabs_curr'><a  id=\"11881_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11881' ><a  id=\"11881_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_11881'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Dada a sucess\u00e3o de termo geral ${d_n} = \\frac{{3n &#8211; 5}}{{n + 2}}$, prove que a sucess\u00e3o \u00e9 limitada.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11881' onClick='GTTabs_show(1,11881)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11881'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Dada a sucess\u00e3o de termo geral ${d_n} = \\frac{{3n &#8211; 5}}{{n + 2}}$, prove que a sucess\u00e3o \u00e9 limitada.<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<p>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{{d_{n + 1}} &#8211; {d_n}}&amp; = &amp;{\\frac{{3\\left( {n + 1} \\right) &#8211; 5}}{{\\left( {n + 1} \\right) + 2}} &#8211; \\frac{{3n &#8211; 5}}{{n + 2}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{3n &#8211; 2}}{{n + 3}} &#8211; \\frac{{3n &#8211; 5}}{{n + 2}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{3{n^2} + 6n &#8211; 2n &#8211; 4 &#8211; 3{n^2} &#8211; 9n + 5n + 15}}{{\\left( {n + 2} \\right)\\left( {n + 3} \\right)}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{11}}{{\\left( {n + 2} \\right)\\left( {n + 3} \\right)}}}<br \/>\n\\end{array}\\]<\/p>\n<p>Como ${d_{n + 1}} &#8211; {d_n} &gt; 0,\\forall n \\in \\mathbb{N}$, ent\u00e3o a sucess\u00e3o $\\left( {{d_n}} \\right)$ \u00e9 estritamente crescente.<\/p>\n<p>\u00ad<\/p>\n<p>Assim, o seu primeiro termo, ${d_1} = \\frac{{3 \\times 1 &#8211; 5}}{{1 + 2}} =\u00a0 &#8211; \\frac{2}{3}$,\u00a0\u00e9\u00a0m\u00ednimo do conjunto dos seus termos, pelo que ser\u00e1: $\\boxed{ &#8211; \\frac{2}{3} \\leqslant {d_n},\\forall n \\in \\mathbb{N}}$ (1).<\/p>\n<p>Por outro lado, temos: \\[\\begin{array}{*{20}{l}}<br \/>\n{{d_n}}&amp; = &amp;{\\frac{{3n &#8211; 5}}{{n + 2}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{3\\left( {n + 2} \\right) &#8211; 11}}{{n + 2}}} \\\\<br \/>\n{}&amp; = &amp;{\\underbrace {3 &#8211; \\underbrace {\\frac{{11}}{{n + 2}}}_{ &gt; 0,\\forall n \\in \\mathbb{N}}}_{ &lt; 3,\\forall n \\in \\mathbb{N}}}<br \/>\n\\end{array}\\]<\/p>\n<p>Logo, $\\boxed{{d_n} &lt; 3,\\forall n \\in \\mathbb{N}}$ (2).<\/p>\n<p>\u00ad<\/p>\n<p>Assim, tendo em considera\u00e7\u00e3o (1)\u00a0e (2), resulta que $ &#8211; \\frac{2}{3} \\leqslant {d_n} &lt; 3,\\forall n \\in \\mathbb{N}$.<\/p>\n<p>Logo, a sucess\u00e3o $\\left( {{d_n}} \\right)$ \u00e9 limitada, pois \u00e9 limitado o conjunto dos seus termos.<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11881' onClick='GTTabs_show(0,11881)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Dada a sucess\u00e3o de termo geral ${d_n} = \\frac{{3n &#8211; 5}}{{n + 2}}$, prove que a sucess\u00e3o \u00e9 limitada. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19431,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,372],"tags":[422,373,431],"series":[],"class_list":["post-11881","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-sucessoes-reais","tag-11-o-ano","tag-sucessao-limitada","tag-sucessoes-reais"],"views":20971,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/05\/Uma_sucessao_d.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11881","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=11881"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11881\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19431"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11881"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11881"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11881"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11881"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}