{"id":235070,"date":"2022-07-07T17:56:53","date_gmt":"2022-07-07T15:56:53","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=235070"},"modified":"2025-09-29T11:31:08","modified_gmt":"2025-09-29T09:31:08","slug":"convergence-dune-suite","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/convergence-dune-suite\/","title":{"rendered":"Convergence d\u2019une suite : Exercice corrig\u00e9"},"content":{"rendered":"\n

Vous travaillez actuellement sur la convergence d\u2019une suite<\/strong> ? Explorez cet article pour comprendre et ma\u00eetriser cette notion fondamentale en math\u00e9matiques, gr\u00e2ce \u00e0 des m\u00e9thodologies adapt\u00e9es et des exercices corrig\u00e9s.<\/p>\n

Tu peines toujours \u00e0 trouver la limite de tes suites ? \ud83c\udf93 Perfectionne ta technique avec un cours particulier de maths<\/strong><\/a>, sp\u00e9cialement con\u00e7u pour les \u00e9tudiants.<\/p>\n

\u00a0<\/p>\n\n\n\n

Exercice 1 : Convergence d’une suite<\/h2>\n\n\n\n

\u23f0 Dur\u00e9e : 20 min<\/p>\n\n\n\n

\ud83d\udcaa Difficult\u00e9 : niveau 1\/3<\/p>\n\n\n\n

<\/p>\n\u00c9tudier la convergence et calculer, si possible, la limite des suites suivantes :
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<\/p>\n1. \"a_n = \sqrt{n^2 + n +1} - n\" ;\n<\/div>\n\n\n\n

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<\/p>\n2. \"b_n = \left\lfloor 1 + \dfrac{\left( - 1 \right)^n}{n} \right\rfloor\" ;\n<\/div>\n\n\n\n

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<\/p>\n3. \"c_n = \dfrac{n \cos \left( n \right)}{n^2+1}\" ;\n<\/div>\n<\/div>\n\n\n\n

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<\/p>\n4. \"d_n = \sin \left( \dfrac{n \pi}{6} \right)\" ;\n<\/div>\n\n\n\n

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<\/p>\n5. \"e_n = \sqrt[n]{2 - \cos \left( n^2 \right)}\" ;\n<\/div>\n\n\n\n

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<\/p>\n6. \"f_n = \dfrac{n^3+2^n}{3^n}\" ;\n<\/div>\n<\/div>\n\n\n\n

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<\/p>\n7. \"g_n = n \tan \left( \dfrac1n \right)\" ;\n<\/div>\n\n\n\n

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<\/p>\n8. \"h_n = \left( - 1 \right)^{\left( - 1 \right)^n}\" ;\n<\/div>\n\n\n\n

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<\/p>\n9. \"\displaystyle{i_n = \sum_{k=1}^n \dfrac{k}{k+n}}\".\n<\/div>\n<\/div>\n\n\n\n

Corrig\u00e9 de l’exercice 1 : Convergence d’une suite<\/h2>\n\n\n\n

<\/p>\n1. On a\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}, \quad \sqrt{n^2 + n +1} - n = \dfrac{\sqrt{n^2 + n +1}^2 - n^2}{\sqrt{n^2 + n +1} + n} = \dfrac{n + 1}{\sqrt{n^2 + n +1} + n}.\]\"<\/p>
\nOn factorise par le terme le plus grand :\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad \dfrac{n + 1}{\sqrt{n^2 + n +1} + n} = \dfrac{n \left( 1 + \dfrac1n \right)}{n \left( \sqrt{1 + \dfrac1n + \dfrac{1}{n^2}} + 1 \right) } = \dfrac{1 + \dfrac1n}{\sqrt{1 + \dfrac1n + \dfrac{1}{n^2}} + 1}.\]\"<\/p>
\nComme \"\lim\limits_{n \to + \infty} \left( 1 + \dfrac1n \right) = 1\" et \"\lim\limits_{n \to + \infty} \left( \sqrt{1 + \dfrac1n + \dfrac{1}{n^2}} + 1 \right) = 2\", on en d\u00e9duit que \n

  <\/span>   <\/span>\"\[\lim_{n \to + \infty} a_n = \dfrac12.\]\"<\/p>
\n2. On a \"b_{2n} = \left\lfloor 1+ \dfrac{1}{2n} \right\rfloor = 1 \xrightarrow[n \to + \infty]{} 1\" et \"b_{2n+1} = \left\lfloor 1 - \dfrac{1}{2n+1} \right\rfloor= 0 \xrightarrow[n \to + \infty]{} 0\".
\nLa suite \"\left( b_n \right)_{n \in \N}\" diverge.
\n3. On a \n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad \dfrac{n \cos \left( n \right)}{n^2+1} = \cos \left( n \right) \times \dfrac{1}{n \left( 1 + \dfrac1n \right)}.\]\"<\/p>\nLa suite \"\left( \cos \left( n \right) \right)_{n \in \mathbb{N}^*}\" est born\u00e9e, la suite \"\left( \dfrac{1}{n \left( 1 + \dfrac1n \right)} \right)_{n \in \mathbb{N}^*}\" converge vers \"0\", donc la suite \"\left( c_n \right)_{n \in \mathbb{N}^*}\" converge vers \"0\".
\n4. On a \"d_{12n} = \sin \left( 2 n \pi \right) = 0 \underset{n \to + \infty}{\longrightarrow} 0\" et \"d_{12n + 3} = \sin \left( 2 n \pi + \dfrac{\pi}{2} \right) = 1 \underset{n \to + \infty}{\longrightarrow} 1\".
\nOn en d\u00e9duit que la suite \"\left( d_n \right)_{n \in \mathbb{N}}\" diverge.
\n5. On a :\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad \sqrt[n]{2 - \cos \left( n^2 \right)} = \exp \left( \dfrac1n \ln \left( 2 - \cos \left( n^2 \right) \right) \right).\]\"<\/p>\nLa suite \"\left( \ln \left( 2 - \cos \left( n^2 \right) \right) \right)_{n \in \mathbb{N}^*}\" est born\u00e9e,
\nla suite \"\left( \dfrac1n \right)_{n \in \mathbb{N}^*}\" converge vers \"0\", donc \"\lim\limits_{n \to + \infty} \dfrac1n \ln \left( 2 - \cos \left( n^2 \right) \right) = 0\".
\nComme \"\lim\limits_{u \to 0} \exp \left( u \right) = 1\", par composition de limites, on a \n

  <\/span>   <\/span>\"\[\lim_{n \to + \infty} e_n = 1.\]\"<\/p>
\n6. On a :\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}, \quad \dfrac{n^3+ 2^n}{3^n} = \dfrac{n^3}{3^n} + \left( \dfrac23 \right)^n.\]\"<\/p>\nPar croissance compar\u00e9e, on a \"\lim\limits_{n \to + \infty} \dfrac{n^3}{3^n} = 0\" et \"\lim\limits_{n \to + \infty} \left( \dfrac23 \right)^n = 0\" car \"\left| \dfrac23 \right| < 1\". On en d\u00e9duit que \"\left( f_n \right)_{n \in \mathbb{N}}\" converge vers \"0\".
\n7. On a :\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad n \tan \left( \dfrac1n \right) = \dfrac{\tan \left( \dfrac1n \right)}{\dfrac1n}.\]\"<\/p>\nComme \"\dfrac1n \underset{n \to + \infty}{\longrightarrow} 0\" et \"\lim\limits_{u \to 0} \dfrac{\tan \left( u \right)}{u} = 1\" (taux d’accroissement), on en d\u00e9duit que la suite \"\left( g_n \right)_{n \in \mathbb{N}^*}\" converge vers \"1\".
\n8. On a \"h_{2n} = - 1\" et \"h_{2n+1} = -1\", la suite \"\left( h_n \right)_{n \in \mathbb{N}}\" est constante \u00e9gale \u00e0 \"- 1\", donc converge vers \"-1\".
\n9. On commence par remarquer que si \"k \in [[1 , n ]]\", alors \"\dfrac{1}{k + n} \ge \dfrac{1}{2n}\", ainsi\n

  <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad i_n \ge \dfrac{1}{2n } \times \sum_{k=1}^n k = \dfrac{n+1}{4} \underset{n \to + \infty}{\longrightarrow} + \infty.\]\"<\/p>\nPar comparaison, on en d\u00e9duit que \"\left( i_n \right)_{n \in \mathbb{N}^*}\" diverge vers \"+ \infty\".\n\n\n\n

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\"livre<\/figure>\n<\/div>\n\n\n\n
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Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/em>(\u00e9ditions Vuibert, juin 2021) <\/em>\u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n<\/div>\n<\/div>\n\n\n\n

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\n 3.1\/5 - (12 votes) <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

Vous travaillez actuellement sur la convergence d\u2019une suite ? Explorez cet article pour comprendre et ma\u00eetriser cette notion (…)<\/p>\n","protected":false},"author":158,"featured_media":164737,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-235070","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235070","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=235070"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235070\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/164737"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=235070"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=235070"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=235070"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}