{"id":17325,"date":"2016-08-16T00:35:42","date_gmt":"2016-08-16T00:35:42","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=17325"},"modified":"2016-08-16T00:35:42","modified_gmt":"2016-08-16T00:35:42","slug":"calculus-convergence-or-divergence-%cf%832%e2%88%9en-1n2-1-0-5","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/calculus-convergence-or-divergence-%cf%832%e2%88%9en-1n2-1-0-5\/","title":{"rendered":"Calculus: convergence (or divergence?): \u03a32<\/sub>\u221e<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>)"},"content":{"rendered":"

The tutor checks an infinite series for convergence.<\/h1>\n

Number 22, page 545, of Larson and Hostetler1<\/sup> asks about the convergence of the series<\/p>\n

Σ2<\/sub>∞<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>)<\/p>\n

Solution:<\/p>\n

First, we realize that, for n>1, n2<\/sup>-1 > (n-1)2<\/sup> Therefore,<\/p>\n

Σ2<\/sub>∞<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>) < Σ2<\/sub>∞<\/sup>(n-1<\/sup>((n-1)2<\/sup>)-0.5<\/sup>)=Σ2<\/sub>∞<\/sup>(n(n-1))-1<\/sup><\/p>\n

In turn, <\/p>\n

Σ2<\/sub>∞<\/sup>(n(n-1))-1<\/sup><Σ2<\/sub>∞<\/sup>(n-1)-2<\/sup>=Σ1<\/sub>∞<\/sup>(n)-2<\/sup><\/p>\n

Therefore, <\/p>\n

Σ2<\/sub>∞<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>) < Σ1<\/sub>∞<\/sup>(n)-2<\/sup><\/p>\n

We know that Σ1<\/sub>∞<\/sup>(n)-2<\/sup> converges; it’s a p-series with p > 1. Therefore, being less than Σ1<\/sub>∞<\/sup>(n)-2<\/sup>, the series in question <\/p>\n

Σ2<\/sub>∞<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>)<\/p>\n

must converge also (by Limit Comparison Test).<\/p>\n

HTH:)<\/p>\n

1<\/sup>Larson, Roland E. and Robert P. Hostetler. Calculus<\/u>, third ed.
Toronto: D C Heath and Company, 1989.<\/p>\n

Jack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"

The tutor checks an infinite series for convergence. Number 22, page 545, of Larson and Hostetler1 asks about the convergence of the series Σ2∞(n-1(n2-1)-0.5) Solution: First, we realize that, for n>1, n2-1 > (n-1)2 Therefore, Σ2∞(n-1(n2-1)-0.5) < Σ2∞(n-1((n-1)2)-0.5)=Σ2∞(n(n-1))-1 In turn, …<\/p>\n

Calculus: convergence (or divergence?): \u03a32<\/sub>\u221e<\/sup>(n-1<\/sup>(n2<\/sup>-1)-0.5<\/sup>)<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[234,3],"tags":[1817,1818,1816],"class_list":["post-17325","post","type-post","status-publish","format-standard","hentry","category-calculus","category-math","tag-limit-comparison-test","tag-p-series","tag-series-convergence"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=17325"}],"version-history":[{"count":20,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17325\/revisions"}],"predecessor-version":[{"id":17345,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17325\/revisions\/17345"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=17325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=17325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=17325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}