{"id":243899,"date":"2025-07-04T17:57:33","date_gmt":"2025-07-04T17:57:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=243899"},"modified":"2025-07-04T17:57:36","modified_gmt":"2025-07-04T17:57:36","slug":"limit-inferior-and-limit-superior-we-define-the-limit-inferior-and-the-limit-superior-of-a-sequence-as-follows","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/limit-inferior-and-limit-superior-we-define-the-limit-inferior-and-the-limit-superior-of-a-sequence-as-follows\/","title":{"rendered":"Limit Inferior and Limit Superior We define the limit inferior and the limit superior of a sequence as follows"},"content":{"rendered":"\n<p>Limit Inferior and Limit Superior We define the limit inferior and the limit superior of a sequence as follows: lim inf an = lim inf{ak | k\u00e2\u2030\u00a5n} = 0+ (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) lim sup an = lim sup{ak | k\u00e2\u2030\u00a5n} = \u00e2\u02c6\u017e (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) Let {an} be a bounded sequence. Show that lim inf an and lim sup an exist and are in \u00e2\u201e\u009d. (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) Let {an} be an unbounded sequence. Show that either lim inf an = -\u00e2\u02c6\u017e or lim sup an = \u00e2\u02c6\u017e (or possibly both). Let {an}, {bn} be two sequences. Show that lim inf (an + bn) \u00e2\u2030\u00a4 lim inf an + lim sup bn \u00e2\u2030\u00a4 lim sup an + lim sup bn (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) Furthermore, find a pair of sequences for which the second inequality is strict.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Definitions:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Limit Inferior<\/strong> (lim inf) of a sequence {an}\\{a_n\\}{an\u200b}: lim\u2009inf\u2061n\u2192\u221ean=lim\u2061n\u2192\u221e(inf\u2061{ak\u2223k\u2265n})\\liminf_{n \\to \\infty} a_n = \\lim_{n \\to \\infty} \\left( \\inf \\{a_k | k \\geq n\\} \\right)n\u2192\u221eliminf\u200ban\u200b=n\u2192\u221elim\u200b(inf{ak\u200b\u2223k\u2265n}) This is the greatest lower bound of the tail of the sequence as n\u2192\u221en \\to \\inftyn\u2192\u221e. Intuitively, it represents the lowest value the sequence &#8220;settles&#8221; on in the long run.<\/li>\n\n\n\n<li><strong>Limit Superior<\/strong> (lim sup) of a sequence {an}\\{a_n\\}{an\u200b}: lim\u2009sup\u2061n\u2192\u221ean=lim\u2061n\u2192\u221e(sup\u2061{ak\u2223k\u2265n})\\limsup_{n \\to \\infty} a_n = \\lim_{n \\to \\infty} \\left( \\sup \\{a_k | k \\geq n\\} \\right)n\u2192\u221elimsup\u200ban\u200b=n\u2192\u221elim\u200b(sup{ak\u200b\u2223k\u2265n}) This is the least upper bound of the tail of the sequence as n\u2192\u221en \\to \\inftyn\u2192\u221e. Intuitively, it represents the highest value the sequence &#8220;settles&#8221; on in the long run.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Bounded Sequence:<\/h3>\n\n\n\n<p>Let {an}\\{a_n\\}{an\u200b} be a <strong>bounded<\/strong> sequence. This means that there exist constants MMM and mmm such that:m\u2264an\u2264Mfor&nbsp;alln.m \\leq a_n \\leq M \\quad \\text{for all} \\quad n.m\u2264an\u200b\u2264Mfor&nbsp;alln.<\/p>\n\n\n\n<p>We need to show that both lim\u2009inf\u2061an\\liminf a_nliminfan\u200b and lim\u2009sup\u2061an\\limsup a_nlimsupan\u200b exist and are real numbers.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Existence:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For each nnn, we can consider the set {ak\u2223k\u2265n}\\{ a_k \\mid k \\geq n \\}{ak\u200b\u2223k\u2265n}, which is a subset of the real numbers and is bounded because ana_nan\u200b is bounded.<\/li>\n\n\n\n<li>The <strong>infimum<\/strong> (greatest lower bound) and <strong>supremum<\/strong> (least upper bound) of this set will be finite because the set of real numbers is complete and the sequence is bounded.<\/li>\n\n\n\n<li>Since the infimum and supremum are non-decreasing and non-increasing respectively as nnn increases, the limits: lim\u2061n\u2192\u221einf\u2061{ak\u2223k\u2265n}andlim\u2061n\u2192\u221esup\u2061{ak\u2223k\u2265n}\\lim_{n \\to \\infty} \\inf \\{a_k \\mid k \\geq n\\} \\quad \\text{and} \\quad \\lim_{n \\to \\infty} \\sup \\{a_k \\mid k \\geq n\\}n\u2192\u221elim\u200binf{ak\u200b\u2223k\u2265n}andn\u2192\u221elim\u200bsup{ak\u200b\u2223k\u2265n} exist by the monotone convergence theorem. Therefore, both the limit inferior and limit superior are well-defined and finite, and they are real numbers.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Unbounded Sequence:<\/h3>\n\n\n\n<p>Let {an}\\{a_n\\}{an\u200b} be an <strong>unbounded<\/strong> sequence. This means that the sequence does not remain within any finite bounds, so either {an}\\{a_n\\}{an\u200b} tends to infinity or negative infinity at some points.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Proof:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If {an}\\{a_n\\}{an\u200b} is <strong>unbounded above<\/strong>, this implies there exists a subsequence where an\u2192+\u221ea_n \\to +\\inftyan\u200b\u2192+\u221e, hence: lim\u2009sup\u2061n\u2192\u221ean=\u221e.\\limsup_{n \\to \\infty} a_n = \\infty.n\u2192\u221elimsup\u200ban\u200b=\u221e.<\/li>\n\n\n\n<li>If {an}\\{a_n\\}{an\u200b} is <strong>unbounded below<\/strong>, this implies there exists a subsequence where an\u2192\u2212\u221ea_n \\to -\\inftyan\u200b\u2192\u2212\u221e, hence: lim\u2009inf\u2061n\u2192\u221ean=\u2212\u221e.\\liminf_{n \\to \\infty} a_n = -\\infty.n\u2192\u221eliminf\u200ban\u200b=\u2212\u221e.<\/li>\n<\/ul>\n\n\n\n<p>Thus, if a sequence is unbounded, we can have either lim\u2009inf\u2061an=\u2212\u221e\\liminf a_n = -\\inftyliminfan\u200b=\u2212\u221e or lim\u2009sup\u2061an=\u221e\\limsup a_n = \\inftylimsupan\u200b=\u221e, or both.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Inequalities involving lim inf and lim sup:<\/h3>\n\n\n\n<p>Now, consider two sequences {an}\\{a_n\\}{an\u200b} and {bn}\\{b_n\\}{bn\u200b}. We want to show the following inequalities:lim\u2009inf\u2061n\u2192\u221e(an+bn)\u2264lim\u2009inf\u2061n\u2192\u221ean+lim\u2009sup\u2061n\u2192\u221ebn\u2264lim\u2009sup\u2061n\u2192\u221e(an+bn).\\liminf_{n \\to \\infty} (a_n + b_n) \\leq \\liminf_{n \\to \\infty} a_n + \\limsup_{n \\to \\infty} b_n \\leq \\limsup_{n \\to \\infty} (a_n + b_n).n\u2192\u221eliminf\u200b(an\u200b+bn\u200b)\u2264n\u2192\u221eliminf\u200ban\u200b+n\u2192\u221elimsup\u200bbn\u200b\u2264n\u2192\u221elimsup\u200b(an\u200b+bn\u200b).<\/p>\n\n\n\n<p>To show this:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>First Inequality<\/strong>: lim\u2009inf\u2061(an+bn)\u2264lim\u2009inf\u2061an+lim\u2009sup\u2061bn\\liminf (a_n + b_n) \\leq \\liminf a_n + \\limsup b_nliminf(an\u200b+bn\u200b)\u2264liminfan\u200b+limsupbn\u200b The sequence an+bna_n + b_nan\u200b+bn\u200b is formed by adding the sequences ana_nan\u200b and bnb_nbn\u200b. By the definition of lim inf, the limit inferior of the sum is at most the sum of the limit inferior of ana_nan\u200b and the limit superior of bnb_nbn\u200b. The intuition is that the infimum of a sum is less than or equal to the sum of the infimum of one sequence and the supremum of the other.<\/li>\n\n\n\n<li><strong>Second Inequality<\/strong>: lim\u2009sup\u2061(an+bn)\u2265lim\u2009sup\u2061an+lim\u2009sup\u2061bn\\limsup (a_n + b_n) \\geq \\limsup a_n + \\limsup b_nlimsup(an\u200b+bn\u200b)\u2265limsupan\u200b+limsupbn\u200b Similarly, the supremum of a sum is at least the sum of the suprema of the individual sequences.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Finding Strict Inequality:<\/h3>\n\n\n\n<p>Consider the following example:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Let an=(\u22121)na_n = (-1)^nan\u200b=(\u22121)n and bn=(\u22121)n+1b_n = (-1)^{n+1}bn\u200b=(\u22121)n+1.\n<ul class=\"wp-block-list\">\n<li>For an=(\u22121)na_n = (-1)^nan\u200b=(\u22121)n, the lim inf and lim sup are: lim\u2009inf\u2061an=\u22121,lim\u2009sup\u2061an=1.\\liminf a_n = -1, \\quad \\limsup a_n = 1.liminfan\u200b=\u22121,limsupan\u200b=1.<\/li>\n\n\n\n<li>For bn=(\u22121)n+1b_n = (-1)^{n+1}bn\u200b=(\u22121)n+1, the lim inf and lim sup are: lim\u2009inf\u2061bn=\u22121,lim\u2009sup\u2061bn=1.\\liminf b_n = -1, \\quad \\limsup b_n = 1.liminfbn\u200b=\u22121,limsupbn\u200b=1.<\/li>\n\n\n\n<li>Now, consider the sum: an+bn=(\u22121)n+(\u22121)n+1.a_n + b_n = (-1)^n + (-1)^{n+1}.an\u200b+bn\u200b=(\u22121)n+(\u22121)n+1. This sequence alternates between 0 and -2. Thus: lim\u2009inf\u2061(an+bn)=\u22122,lim\u2009sup\u2061(an+bn)=0.\\liminf (a_n + b_n) = -2, \\quad \\limsup (a_n + b_n) = 0.liminf(an\u200b+bn\u200b)=\u22122,limsup(an\u200b+bn\u200b)=0.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>In this case:lim\u2009inf\u2061(an+bn)=\u22122,lim\u2009inf\u2061an+lim\u2009sup\u2061bn=\u22121+1=0.\\liminf (a_n + b_n) = -2, \\quad \\liminf a_n + \\limsup b_n = -1 + 1 = 0.liminf(an\u200b+bn\u200b)=\u22122,liminfan\u200b+limsupbn\u200b=\u22121+1=0.<\/p>\n\n\n\n<p>Here, the second inequality is strict:lim\u2009inf\u2061(an+bn)&lt;lim\u2009inf\u2061an+lim\u2009sup\u2061bn.\\liminf (a_n + b_n) &lt; \\liminf a_n + \\limsup b_n.liminf(an\u200b+bn\u200b)&lt;liminfan\u200b+limsupbn\u200b.<\/p>\n\n\n\n<p>This demonstrates that the second inequality can indeed be strict in certain cases.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-372.jpeg\" alt=\"\" class=\"wp-image-243914\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Limit Inferior and Limit Superior We define the limit inferior and the limit superior of a sequence as follows: lim inf an = lim inf{ak | k\u00e2\u2030\u00a5n} = 0+ (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) lim sup an = lim sup{ak | k\u00e2\u2030\u00a5n} = \u00e2\u02c6\u017e (n\u00e2\u2020\u2019\u00e2\u02c6\u017e) Let {an} be a bounded sequence. Show that lim inf an and lim sup [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-243899","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243899","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=243899"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/243899\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=243899"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=243899"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=243899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}