{"id":948,"date":"2025-06-20T17:24:31","date_gmt":"2025-06-20T17:24:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=948"},"modified":"2025-09-05T15:20:05","modified_gmt":"2025-09-05T15:20:05","slug":"taylor-and-maclaurin-series-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-learn-it-4\/","title":{"raw":"Taylor and Maclaurin Series: Learn It 4","rendered":"Taylor and Maclaurin Series: Learn It 4"},"content":{"raw":"

Convergence of Taylor Series<\/h2>\r\nWe now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.\r\n\r\n
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Find the Taylor series for [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] at [latex]x=1[\/latex]. Determine the interval of convergence.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558999\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558999\"]\r\n

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For [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex], the values of the function and its first four derivatives at [latex]x=1[\/latex] are<\/p>\r\n\r\n

[latex]\\begin{array}{cccccccc}\\hfill f\\left(x\\right)& =\\hfill & \\frac{1}{x}\\hfill & & & \\hfill f\\left(1\\right)& =\\hfill & 1\\hfill \\\\ \\hfill {f}^{\\prime }\\left(x\\right)& =\\hfill & -\\frac{1}{{x}^{2}}\\hfill & & & \\hfill {f}^{\\prime }\\left(1\\right)& =\\hfill & -1\\hfill \\\\ \\hfill f^{\\prime\\prime}\\left(x\\right)& =\\hfill & \\frac{2}{{x}^{3}}\\hfill & & & \\hfill f^{\\prime\\prime}\\left(1\\right)& =\\hfill & 2\\text{!}\\hfill \\\\ \\hfill f^{\\prime\\prime\\prime}\\left(x\\right)& =\\hfill & -\\frac{3\\cdot 2}{{x}^{4}}\\hfill & & & \\hfill f^{\\prime\\prime\\prime}\\left(1\\right)& =\\hfill & -3\\text{!}\\hfill \\\\ \\hfill {f}^{\\left(4\\right)}\\left(x\\right)& =\\hfill & \\frac{4\\cdot 3\\cdot 2}{{x}^{5}}\\hfill & & & \\hfill {f}^{\\left(4\\right)}\\left(1\\right)& =\\hfill & 4\\text{!.}\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

That is, we have [latex]{f}^{\\left(n\\right)}\\left(1\\right)={\\left(-1\\right)}^{n}n\\text{!}[\/latex] for all [latex]n\\ge 0[\/latex]. Therefore, the Taylor series for [latex]f[\/latex] at [latex]x=1[\/latex] is given by<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(1\\right)}{n\\text{!}}{\\left(x - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex].<\/div>\r\n \r\n

To find the interval of convergence, we use the ratio test. We find that<\/p>\r\n\r\n

[latex]\\frac{|{a}_{n+1}|}{|{a}_{n}|}=\\frac{|{\\left(-1\\right)}^{n+1}{\\left(x - 1\\right)}^{n+1}|}{|{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}|}=|x - 1|[\/latex].<\/div>\r\n \r\n

Thus, the series converges if [latex]|x - 1|<1[\/latex]. That is, the series converges for [latex]0<x<2[\/latex]. Next, we need to check the endpoints. At [latex]x=2[\/latex], we see that<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(2 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}[\/latex]<\/div>\r\n \r\n

diverges by the divergence test. Similarly, at [latex]x=0[\/latex],<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(0 - 1\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }1[\/latex]<\/div>\r\n \r\n

diverges. Therefore, the interval of convergence is [latex]\\left(0,2\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Watch the following video to see the worked solution to the example above.