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| {{Calculus |Series}}
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| In [[mathematics]], an '''alternating series''' is an [[infinite series]] of the form
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| :<math>\sum_{n=0}^\infty (-1)^n\,a_n</math> or <math>\sum_{n=1}^\infty (-1)^{n-1}\,a_n</math>
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| with ''a<sub>n</sub>'' > 0 for all ''n''. The signs of the general terms alternate between positive and negative. Like any series, an alternating [[Convergent series|series converges]] if and only if the associated sequence of partial sums [[Limit of a sequence|converges]].
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| == Alternating series test ==
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| {{main|alternating series test}}
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| The theorem known as "Leibniz Test" or the [[alternating series test]] tells us that an alternating series will converge if the terms ''a<sub>n</sub>'' converge to 0 [[monotonic function|monotonically]].
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| Proof: Suppose the sequence <math>a_n</math> converges to zero and is monotone decreasing. If <math>m</math> is odd and <math>m<n</math>, we obtain the estimate <math>S_m - S_n < a_{m}</math> via the following calculation:
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| : <math>
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| \begin{align}
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| S_m - S_n & =
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| \sum_{k=0}^m(-1)^k\,a_k\,-\,\sum_{k=0}^n\,(-1)^k\,a_k\ = \sum_{k=m+1}^n\,(-1)^k\,a_k \\
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| & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\
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| & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) - (a_{m+4}-a_{m+5}) -\cdots-a_n \le a_{m+1}\le a_{m} [a_{n} decreasing].
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| \end{align}
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| </math>
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| Since <math>a_n</math> is monotonically decreasing, the terms <math>-(a_m - a_{m+1})</math> are negative. Thus, we have the final inequality <math>S_m - S_n \le a_{m}</math> .Similarly it can be shown that <math>-a_{m}\le S_m - S_n </math>. Since <math>a_{m}</math> converges to <math>0</math>, our partial sums <math>S_m</math> form a [[Cauchy sequence]] (i.e. the series satisfies the [[Cauchy convergence criterion for series]]) and therefore converge. The argument for <math>m</math> even is similar.
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| == Approximating sums ==
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| The estimate above does not depend on <math>n</math>. So, if <math>a_n</math> is approaching 0 monotonically, the estimate provides an [[error bound]] for approximating infinite sums by partial sums:
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| : <math>|\sum_{k=0}^\infty(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k|\le |a_{m+1}|.</math>
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| == Absolute convergence ==
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| A series <math>\sum a_n</math> [[absolute convergence|converges absolutely]] if the series <math>\sum |a_n|</math> converges.
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| Theorem: Absolutely convergent series are convergent.
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| Proof: Suppose <math>\sum a_n</math> is absolutely convergent. Then, <math>\sum |a_n|</math> is convergent and it follows that <math>\sum 2|a_n|</math> converges as well. Since <math> 0 \leq a_n + |a_n| \leq 2|a_n|</math>, the series <math>\sum (a_n + |a_n|)</math> converges by the [[comparison test]]. Therefore, the series <math>\sum a_n</math> converges as the difference of two convergent series <math>\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|</math>.
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| == Conditional convergence ==
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| A series is [[Conditional convergence|conditionally convergent]] if it converges but does not converge absolutely.
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| For example, the [[harmonic series (mathematics)|harmonic series]]
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| :<math>\sum_{n=1}^\infty \frac{1}{n},\! </math>
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| diverges, while the alternating version
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| :<math>\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n},\! </math>
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| converges by the [[Alternating_series#Alternating_series_test|alternating series test]].
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| == Rearrangements ==
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| For any series, we can create a new series by rearranging the order of summation. A series is [[Series_(mathematics)#Unconditionally_convergent_series|unconditionally convergent]] if any rearrangement creates a series with the same convergence as the original series. [[Absolute_convergence#Rearrangements_and_unconditional_convergence|Absolutely convergent series are unconditionally convergent]]. But the [[Riemann series theorem]] states that conditionally convergent series can be rearranged to create arbitrary convergence.<ref>{{cite journal |last1=Mallik |first1=AK|year=2007 |title=Curious Consequences of Simple Sequences |journal=Resonance |volume=12 |issue=1 |pages=23–37 |url=http://www.springerlink.com/index/D65WX2N5384880LV.pdf}}</ref> The general principle is that addition of infinite sums is only commutative for absolutely convergent series.
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| For example, this [[Mathematical_fallacy#Associative_law|false proof that 1=0]] exploits the failure of associativity for infinite sums.
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| As another example, [[Natural_logarithm#Derivative.2C_Taylor_series|we know that]]
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| :<math>\ln(2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots.</math>
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| But, since the series does not converge absolutely, we can rearrange the terms to obtain a series for <math>\frac{1}{2}\ln(2)</math>:
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| :<math>
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| \begin{align}
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| & {} \quad \left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots \\[8pt]
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| & = \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots \\[8pt]
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| & = \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots\right)= \frac{1}{2} \ln(2).
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| \end{align}
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| </math>
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| Another valid example of alternating series is the following
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| :<math>
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| \sum_{k=0}^{\infty}
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| \frac{(-1)^{k}}{\sqrt{k+1}}=1-\frac{1}{\sqrt{2}}
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| +\frac{1}{\sqrt{3}}
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| -\frac{1}{\sqrt{4}}
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| +\frac{1}{\sqrt{5}}
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| \cdots=-(\sqrt{2}
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| -1)\zeta(\frac{1}{2})\approx0.6048986434....
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| </math>
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| == Series acceleration ==
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| In practice, the numerical summation of an alternating series may be sped up using any one of a variety of [[series acceleration]] techniques. One of the oldest techniques is that of [[Euler summation]], and there are many modern techniques that can offer even more rapid convergence.
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| ==See also==
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| * [[Nörlund–Rice integral]]
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| * [[Series (mathematics)]]
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| ==Notes==
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| {{reflist}}
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| ==References==
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| *{{MathWorld|title=Alternating Series|urlname=AlternatingSeries}}
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| {{DEFAULTSORT:Alternating Series}}
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| [[Category:Calculus]]
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| [[Category:Mathematical series]]
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| [[Category:Real analysis]]
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The author's title is Milford Wyant. To perform domino is 1 of the things he loves most. Invoicing is how I make a living and I don't think I'll alter it anytime soon. He's always cherished living in Delaware and his mothers and fathers live close by. See what's new on her web site here: https://audanika.zendesk.com/entries/46057239-Why-Everything-You-Know-About-Nya-Online-Casino-Is-A-Lie
Feel free to visit my website; Nya svenska språket online casino 2014 (Gå På dette nettstedet)