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Concept Version 10
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Alternating Series

An alternating series is an infinite series of the form ∑n=0∞(−1)nan\sum_{n=0}^\infty (-1)^n\,a_n∑​n=0​∞​​(−1)​n​​a​n​​ or ∑n=0∞(−1)n−1an\sum_{n=0}^\infty (-1)^{n-1}\,a_n∑​n=0​∞​​(−1)​n−1​​a​n​​ with an>0a_n > 0a​n​​>0 for all nnn.

Learning Objective

  • Describe the properties of an alternating series and the requirements for one to converge


Key Points

    • The theorem known as the "Leibniz Test," or the alternating series test, tells us that an alternating series will converge if the terms ana_na​n​​ converge to 000 monotonically.
    • The signs of the general terms alternate between positive and negative.
    • The sum ∑n=1∞(−1)n+1n\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}∑​n=1​∞​​​n​​(−1)​n+1​​​​ converges by the alternating series test.

Terms

  • monotone

    property of a function to be either always decreasing or always increasing

  • Cauchy sequence

    a sequence whose elements become arbitrarily close to each other as the sequence progresses


Full Text

An alternating series is an infinite series of the form:

∑n=0∞(−1)nan\displaystyle{\sum_{n=0}^\infty (-1)^n\,a_n}​n=0​∑​∞​​(−1)​n​​a​n​​

or:

∑n=0∞(−1)n−1an\displaystyle{\sum_{n=0}^\infty (-1)^{n-1}\,a_n}​n=0​∑​∞​​(−1)​n−1​​a​n​​

with an>0a_n > 0a​n​​>0 for all nnn. The signs of the general terms alternate between positive and negative. Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.

Alternating Series Test

The theorem known as the "Leibniz Test," or the alternating series test, tells us that an alternating series will converge if the terms ana_na​n​​ converge to 000 monotonically.

Proof: Suppose the sequence ana_na​n​​ converges to 000 and is monotone decreasing. If mmm is odd and Sm−Sn<amS_m - S_n < a_{m}S​m​​−S​n​​<a​m​​ via the following calculation: 

Sm−Sn=∑k=0m(−1)kak−∑k=0n(−1)kak =∑k=m+1n(−1)kak=am+1−am+2+am+3−am+4+⋯+an=am+1−(am+2−am+3)−⋯−an≤am+1≤am[an decreasing].\begin{aligned} S_m - S_n & = \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 \\ & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\ & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) -\cdots-a_n \le a_{m+1}\le a_{m} \\& \quad [a_{n} \text{ decreasing}]. \end{aligned}​S​m​​−S​n​​​​​​​​​=∑​k=0​m​​(−1)​k​​a​k​​−∑​k=0​n​​(−1)​k​​a​k​​ ​=∑​k=m+1​n​​(−1)​k​​a​k​​​=a​m+1​​−a​m+2​​+a​m+3​​−a​m+4​​+⋯+a​n​​​=a​m+1​​−(a​m+2​​−a​m+3​​)−⋯−a​n​​≤a​m+1​​≤a​m​​​[a​n​​ decreasing].​​

Since ana_na​n​​ is monotonically decreasing, the terms are negative. Thus, we have the final inequality Sm−Sn≤amS_m - S_n \le a_{m}S​m​​−S​n​​≤a​m​​. Similarly, it can be shown that, since ama_ma​m​​ converges to 000, Sm−SnS_m - S_nS​m​​−S​n​​ converges to 000 for m,n→∞m, n \rightarrow \inftym,n→∞. Therefore, our partial sum SmS_mS​m​​ converges. (The sequence {Sm}\{ S_m \}{S​m​​} is said to form a Cauchy sequence, meaning that elements of the sequence become arbitrarily close to each other as the sequence progresses.) The argument for mmm even is similar.

Example: 

∑n=1∞(−1)n+1n=1−12+13−14+⋯\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots}​n=1​∑​∞​​​n​​(−1)​n+1​​​​=1−​2​​1​​+​3​​1​​−​4​​1​​+⋯

an=1na_n = \frac1na​n​​=​n​​1​​ converges to 0 monotonically. Therefore, the sum ∑n=1∞(−1)n+1n\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}∑​n=1​∞​​​n​​(−1)​n+1​​​​ converges by the alternating series test.

Alternating Harmonic Series

The first fourteen partial sums of the alternating harmonic series (black line segments) shown converging to the natural logarithm of 2 (red line).

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