{"id":47707,"date":"2025-07-02T15:21:03","date_gmt":"2025-07-02T15:21:03","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47707"},"modified":"2025-07-02T15:21:05","modified_gmt":"2025-07-02T15:21:05","slug":"find-the-interval-of-convergence-of-the-power-series","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-interval-of-convergence-of-the-power-series\/","title":{"rendered":"Find the interval of convergence of the power series."},"content":{"rendered":"\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"519\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-134.png\" alt=\"\" class=\"wp-image-47709\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-134.png 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-134-300x152.png 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-134-768x389.png 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Based on the provided solution for part (a), there appears to be a typo in the problem statement for f(x). The function as written, f(x) = \u03a3 [(-1)^(n+1)(x-2)^n] \/ n, has a radius of convergence of R=1, leading to an interval of (1, 4), which contradicts the given answer of (0, 4].<\/p>\n\n\n\n<p>For the interval to be (0, 4], the center must be a=2 and the radius must be R=2. This occurs if the function is:<br><strong>f(x) = \u03a3 (from n=1 to \u221e) [(-1)^(n+1)(x-2)^n] \/ (n * 2^n)<\/strong><br>We will proceed using this corrected function.<\/p>\n\n\n\n<p><strong>(a) Interval of convergence for f(x)<\/strong><br>The interval of convergence for this corrected function is indeed&nbsp;<strong>(0, 4]<\/strong>. At x=0, the series becomes -\u03a3 1\/n, the divergent harmonic series. At x=4, it becomes \u03a3 (-1)^(n+1)\/n, the convergent alternating harmonic series.<\/p>\n\n\n\n<p><strong>(b) Interval of convergence for f'(x)<\/strong><br>Term-by-term differentiation gives:<br>f'(x) = \u03a3 (from n=1 to \u221e) [(-1)^(n+1) * n(x-2)^(n-1)] \/ (n * 2^n) = \u03a3 (from n=1 to \u221e) [(-1)^(n+1)(x-2)^(n-1)] \/ 2^n<br>The radius of convergence remains R=2, so the open interval is (0, 4). We check the endpoints:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At x=0: The series is \u03a3 [(-1)^(n+1)(-2)^(n-1)] \/ 2^n = \u03a3 [(-1)^(2n)] \/ 2 = \u03a3 1\/2. This diverges by the n-th Term Test.<\/li>\n\n\n\n<li>At x=4: The series is \u03a3 [(-1)^(n+1)(2)^(n-1)] \/ 2^n = \u03a3 (-1)^(n+1) \/ 2. This diverges as its partial sums oscillate.<br>The interval of convergence for f'(x) is\u00a0<strong>(0, 4)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>(c) Interval of convergence for f&#8221;(x)<\/strong><br>Differentiating f'(x) gives:<br>f&#8221;(x) = \u03a3 (from n=2 to \u221e) [(-1)^(n+1)(n-1)(x-2)^(n-2)] \/ 2^n<br>The radius is still R=2. We check the endpoints of (0, 4):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At x=0: The series is \u03a3 [(-1)^(n+1)(n-1)(-2)^(n-2)] \/ 2^n = \u03a3 -(n-1)\/4. This diverges.<\/li>\n\n\n\n<li>At x=4: The series is \u03a3 [(-1)^(n+1)(n-1)(2)^(n-2)] \/ 2^n = \u03a3 (-1)^(n+1)(n-1)\/4. This diverges.<br>The interval of convergence for f&#8221;(x) is\u00a0<strong>(0, 4)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>(d) Interval of convergence for \u222bf(x) dx<\/strong><br>Term-by-term integration gives:<br>\u222bf(x) dx = C + \u03a3 (from n=1 to \u221e) [(-1)^(n+1)(x-2)^(n+1)] \/ (n(n+1) * 2^n)<br>The radius is still R=2. We check the endpoints of (0, 4):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At x=0: The series is \u03a3 [(-1)^(n+1)(-2)^(n+1)] \/ (n(n+1)2^n) = \u03a3 2\/(n(n+1)). This series converges by comparison with the p-series \u03a3 1\/n\u00b2.<\/li>\n\n\n\n<li>At x=4: The series is \u03a3 [(-1)^(n+1)(2)^(n+1)] \/ (n(n+1)2^n) = \u03a3 2(-1)^(n+1)\/(n(n+1)). This series converges absolutely (as shown for x=0), so it converges.<br>The interval of convergence for \u222bf(x) dx is\u00a0<strong>[0, 4]<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Final Answers:<\/strong><br>(a) f(x):&nbsp;<strong>(0, 4]<\/strong><br>(b) f'(x):&nbsp;<strong>(0, 4)<\/strong><br>(c) f&#8221;(x):&nbsp;<strong>(0, 4)<\/strong><br>(d) \u222b f(x) dx:&nbsp;<strong>[0, 4]<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-271.jpeg\" alt=\"\" class=\"wp-image-47714\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-271.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-271-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-271-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-271-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is: Based on the provided solution for part (a), there appears to be a typo in the problem statement for f(x). The function as written, f(x) = \u03a3 [(-1)^(n+1)(x-2)^n] \/ n, has a radius of convergence of R=1, leading to an interval of (1, 4), which contradicts the given answer [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47707","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47707","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47707"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47707\/revisions"}],"predecessor-version":[{"id":47717,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47707\/revisions\/47717"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47707"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47707"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}