{"id":237460,"date":"2022-06-08T17:14:14","date_gmt":"2022-06-08T15:14:14","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=237460"},"modified":"2024-08-28T22:47:07","modified_gmt":"2024-08-28T20:47:07","slug":"convergence-des-series-numeriques","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/convergence-des-series-numeriques\/","title":{"rendered":"Convergence des s\u00e9ries num\u00e9riques : exercice corrig\u00e9"},"content":{"rendered":"\n

En gal\u00e8re sur un exercice de convergence des s\u00e9ries num\u00e9riques<\/strong> ? Obtiens des m\u00e9thodologies compl\u00e8tes et adapt\u00e9es gr\u00e2ce \u00e0 cet article d\u00e9di\u00e9 \u00e0 la notion.\u00a0<\/strong><\/p>\n

Avant ta prochaine interro sur les s\u00e9ries, assure-toi d’\u00eatre au top avec l’aide d’un prof particulier de maths en ligne<\/strong><\/a>\u00a0qui saura te guider vers la r\u00e9ussite. \ud83d\udcda<\/span><\/p>\n\n\n\n

Exercice 1 : Convergence des s\u00e9ries num\u00e9riques <\/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>\n1. Montrer que la s\u00e9rie \"\displaystyle{\sum_{n \ge 0} \frac{1}{n!}}\" converge. On admet que \"\displaystyle{\sum_{n=0}^{+ \infty} \frac{1}{n!} = \mathrm{e}}\".
\n2. Montrer que pour tout polyn\u00f4me \"P \in \mathbb{R} \left[ X \right]\", la s\u00e9rie \"\displaystyle{\sum_{n \ge 0} \frac{P \left( n \right)}{n!}}\" converge. On note \"S \left( P \right)\" sa somme.
\n3. Calculer \"S \left( X + 1 \right)\" et \"S \left( X^2 \right)\".\n\n\n\n

Corrig\u00e9 de l’exercice 1 : Convergence des s\u00e9ries num\u00e9riques<\/h2>\n\n\n\n

<\/p>\n1. Par croissances compar\u00e9es, on a \"\dfrac{1}{n!} \underset{n \to + \infty}{=} o \left( \dfrac{1}{n^2} \right)\". Or, la s\\’erie \"\displaystyle{\sum_{n \ge 1} \frac{1}{n^2}}\" converge (s\u00e9rie de Riemann), par n\u00e9gligeabilit\u00e9, la s\u00e9rie \"\displaystyle{\sum_{n \ge 0} \dfrac{1}{n!}}\" converge.
\n2. Comme un polyn\u00f4me est combinaison lin\u00e9aire de mon\u00f4mes, il suffit de montrer que pour tout mon\u00f4me \"X^k\", la s\u00e9rie \"\displaystyle{\sum_{n \ge 0} \frac{n^k}{n!}}\" converge. Or, \"n^k \underset{n \to + \infty}{\sim} \dfrac{n!}{\left( n - k \right)!}\", ainsi \"\dfrac{n^k}{n!} \underset{n \to + \infty}{\sim} \dfrac{1}{\left( n - k \right)!}\".\n\nOr, la s\u00e9rie \"\displaystyle{\sum_{n \ge k} \frac{1}{\left( n - k \right)!}}\" converge, par \u00e9quivalence des s\u00e9ries \u00e0 termes positifs, la s\u00e9rie \"\displaystyle{\sum_{n \ge 0} \dfrac{n^k}{n!}}\" converge.
\n3. Par relation de Chasles, on a :\n

  <\/span>   <\/span>\"\begin{eqnarray*}<\/p>\n

  • On commence par remarquer que pour tout \"n \in \mathbb{N}\", \"n^2 = n \left( n - 1 \right) + n\". Ainsi :\n

      <\/span>   <\/span>\"\begin{eqnarray*}<\/p> <\/li>\n\n\n\n

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    \"livre<\/a><\/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 <\/a><\/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 5\/5 - (1 vote) <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

    En gal\u00e8re sur un exercice de convergence des s\u00e9ries num\u00e9riques ? Obtiens des m\u00e9thodologies compl\u00e8tes et adapt\u00e9es gr\u00e2ce (…)<\/p>\n","protected":false},"author":158,"featured_media":203453,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-237460","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\/237460","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=237460"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/237460\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/203453"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=237460"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=237460"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=237460"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}