{"id":237080,"date":"2022-06-08T17:14:39","date_gmt":"2022-06-08T15:14:39","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=237080"},"modified":"2025-09-29T11:26:06","modified_gmt":"2025-09-29T09:26:06","slug":"convergence-dune-serie","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/convergence-dune-serie\/","title":{"rendered":"Comment montrer la convergence d’une s\u00e9rie ?"},"content":{"rendered":"\n

Vous \u00e9tudiez actuellement la convergence d’une s\u00e9rie<\/strong> ? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9 \u00e0 la notion Comment montrer la convergence d’une s\u00e9rie ?<\/strong>, gr\u00e2ce \u00e0 une m\u00e9thodologie bien rod\u00e9e, r\u00e9ussi avec brio tes prochains exercices et contr\u00f4les sur cette notion !<\/p>\n

Avant de plonger dans ton prochain contr\u00f4le, assure-toi aussi de ma\u00eetriser la convergence des s\u00e9ries avec l’aide d’un Sherpa expert en math\u00e9matiques par visio<\/strong><\/a>\u00a0d\u00e9di\u00e9 \u00e0 ton succ\u00e8s. \ud83d\udcd8<\/p>\n\n\n\n

M\u00e9thode 1 : Montrer la convergence d’une s\u00e9rie et calculer sa somme<\/h2>\n\n\n\n

<\/p>\nMontrons la convergence et calculons la somme de la s\u00e9rie \"\displaystyle{\sum_{n \ge 1} \mathrm{arctan} \left( \frac{1}{n^2+ n +1} \right)}\".
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  • En utilisant \"\mathrm{arctan} \left( x\right)\underset{x \to 0}{\sim} x\", on en d\u00e9duit que \n

      <\/span>   <\/span>\"\[\mathrm{arctan} \left( \frac{1}{n^2+ n +1} \right) \underset{n \to + \infty}{\sim} \frac{1}{n^2+n+1} \underset{n \to + \infty}{\sim} \frac{1}{n^2}.\]\"<\/p>\n Par \u00e9quivalence avec une s\u00e9rie de Riemann convergente, on en d\u00e9duit que la s\u00e9rie \"\displaystyle{\sum_{n \ge 1} \mathrm{arctan} \left( \frac{1}{n^2+ n +1} \right)}\" converge. <\/li>\n

  • On remarque que pour tout \"n \in \mathbb{N}^*\", \"\mathrm{arctan} \left( \dfrac{1}{n^2+ n +1} \right) = \mathrm{arctan} \left( n +1 \right) - \mathrm{arctan} \left( n \right)\". En effet, \"\tan \left( \mathrm{arctan} \left( \dfrac{1}{n^2+ n +1} \right) \right) = \dfrac{1}{n^2+n+1}\" et \"\tan \left( \mathrm{arctan} \left( n +1 \right) - \mathrm{arctan} \left( n \right) \right) = \dfrac{1}{n^2+n +1}\". Comme \"\mathrm{arctan} \left( \dfrac{1}{n^2+ n +1} \right)\" et \"\mathrm{arctan} \left( n +1 \right) - \mathrm{arctan} \left( n \right)\" sont des \\’el\\’ements de \"\left]- \dfrac{\pi}{2} , \dfrac{\pi}{2} \right[\", on en d\u00e9duit que \n

      <\/span>   <\/span>\"\[\forall n \in \mathbb{N}^*, \quad \mathrm{arctan} \left( \dfrac{1}{n^2+ n +1} \right) = \mathrm{arctan} \left( n +1 \right) - \mathrm{arctan} \left( n \right).\]\"<\/p>\n Soit \"N \in \mathbb{N}^*\". Par t\u00e9lescopage, on a \"\displaystyle{\sum_{n = 1}^N \mathrm{arctan} \left( \frac{1}{n^2+ n +1} \right) = \mathrm{arctan} \left( N +1 \right) - \mathrm{arctan} \left( 1 \right)}\". Comme \"\lim\limits_{N \to + \infty} \left( \mathrm{arctan} \left( N +1 \right) - \mathrm{arctan} \left( 1 \right) \right) = \dfrac{\pi}{4}\", on en d\u00e9duit que \"\displaystyle{\sum_{n = 1}^{+ \infty} \mathrm{arctan} \left( \frac{1}{n^2+ n +1} \right) = \frac{\pi}{4}}\". <\/li>\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) \u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n\n <\/div>\n <\/section>\n<\/div>\n<\/div>\n\n\n

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    \n Tu as aim\u00e9 cet article ?<\/span>\n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

    Vous \u00e9tudiez actuellement la convergence d’une s\u00e9rie ? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9 \u00e0 la notion Comment montrer (…)<\/p>\n","protected":false},"author":158,"featured_media":164834,"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-237080","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\/237080","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=237080"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/237080\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/164834"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=237080"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=237080"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=237080"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}