{"id":1112,"date":"2025-06-30T16:34:47","date_gmt":"2025-06-30T16:34:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1112"},"modified":"2025-08-18T15:17:33","modified_gmt":"2025-08-18T15:17:33","slug":"alternating-series-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/alternating-series-learn-it-2\/","title":{"raw":"Alternating Series: Learn It 2","rendered":"Alternating Series: Learn It 2"},"content":{"raw":"

Remainder of an Alternating Series<\/h2>\r\n

When working with alternating series, we often need to approximate the infinite sum using partial sums. A key question becomes: how accurate is our approximation? The good news is that for alternating series satisfying the alternating series test, we can easily bound the error.<\/p>\r\nConsider an alternating series [latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n[\/latex] that satisfies the conditions of the alternating series test. Let [latex]S[\/latex] represent the exact sum and [latex]{S_k}[\/latex] be the sequence of partial sums.\r\n

From Figure 2, we can see that for any integer [latex]N \\geq 1[\/latex], the remainder [latex]R_N[\/latex] satisfies:<\/p>\r\n\r\n

[latex]|{R}_{N}|=|S-{S}_{N}|\\le |{S}_{N+1}-{S}_{N}|={b}_{n+1}[\/latex].<\/div>\r\n
\r\n

This observation leads to an important theorem.<\/p>\r\n\r\n<\/div>\r\n

\r\n

theorem: remainders in alternating series<\/h3>\r\n

Consider an alternating series of the form:<\/p>\r\n\r\n

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

that satisfies the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of the series and [latex]{S}_{N}[\/latex] denote the [latex]N\\text{th}[\/latex] partial sum. For any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}=S-{S}_{N}[\/latex] satisfies:<\/p>\r\n\r\n

[latex]|{R}_{N}|\\le {b}_{N+1}[\/latex].<\/div>\r\n<\/section>
What This Means:<\/strong> If the alternating series test conditions are met, the error in approximating the infinite series by the [latex]N[\/latex]th partial sum is at most the size of the very next term [latex]b_{N+1}[\/latex]. This makes error estimation remarkably straightforward!\r\n\r\n<\/section>
\r\n
\r\n

Consider the alternating series<\/p>\r\n\r\n

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

Use the remainder estimate to determine a bound on the error [latex]{R}_{10}[\/latex] if we approximate the sum of the series by the partial sum [latex]{S}_{10}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n

\r\n

From the theorem stated above,<\/p>\r\n

[latex]|{R}_{10}|\\le {b}_{11}=\\frac{1}{{11}^{2}}\\approx 0.008265[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

Remainder of an Alternating Series<\/h2>\n

When working with alternating series, we often need to approximate the infinite sum using partial sums. A key question becomes: how accurate is our approximation? The good news is that for alternating series satisfying the alternating series test, we can easily bound the error.<\/p>\n

Consider an alternating series [latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n[\/latex] that satisfies the conditions of the alternating series test. Let [latex]S[\/latex] represent the exact sum and [latex]{S_k}[\/latex] be the sequence of partial sums.<\/p>\n

From Figure 2, we can see that for any integer [latex]N \\geq 1[\/latex], the remainder [latex]R_N[\/latex] satisfies:<\/p>\n

[latex]|{R}_{N}|=|S-{S}_{N}|\\le |{S}_{N+1}-{S}_{N}|={b}_{n+1}[\/latex].<\/div>\n
\n

This observation leads to an important theorem.<\/p>\n<\/div>\n

\n
\n

theorem: remainders in alternating series<\/h3>\n

Consider an alternating series of the form:<\/p>\n

[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/div>\n

 <\/p>\n

that satisfies the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of the series and [latex]{S}_{N}[\/latex] denote the [latex]N\\text{th}[\/latex] partial sum. For any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}=S-{S}_{N}[\/latex] satisfies:<\/p>\n

[latex]|{R}_{N}|\\le {b}_{N+1}[\/latex].<\/div>\n<\/section>\n
What This Means:<\/strong> If the alternating series test conditions are met, the error in approximating the infinite series by the [latex]N[\/latex]th partial sum is at most the size of the very next term [latex]b_{N+1}[\/latex]. This makes error estimation remarkably straightforward!<\/p>\n<\/section>\n
\n
\n

Consider the alternating series<\/p>\n

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

 <\/p>\n

Use the remainder estimate to determine a bound on the error [latex]{R}_{10}[\/latex] if we approximate the sum of the series by the partial sum [latex]{S}_{10}[\/latex].<\/p>\n<\/div>\n