Does The Alternating Harmonic Series Converge

The Alternating HarmonicSeries: Does It Converge?

The alternating harmonic series represents one of the most fascinating and fundamental examples in the study of infinite series, particularly when examining convergence behavior. This series, defined as the sum of the terms ( \frac{(-1)^{n+1}}{n} ) for ( n ) starting from 1, presents a compelling case where partial sums oscillate but ultimately settle on a specific value. Understanding its convergence is crucial not only for mastering calculus but also for appreciating deeper concepts in analysis and real analysis. This article delves into the nature of this series, exploring the rigorous mathematical reasoning behind its convergence.

Introduction The harmonic series, ( \sum_{n=1}^{\infty} \frac{1}{n} ), is a well-known series that diverges. Its partial sums grow without bound, albeit very slowly. The alternating harmonic series modifies this by introducing alternating signs: ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots ). While the harmonic series diverges, the alternating version exhibits a different behavior. The central question is: does the alternating harmonic series converge? The answer, supported by the Alternating Series Test, is a definitive yes. This test provides a clear criterion for convergence when terms decrease monotonically in absolute value and approach zero. For the alternating harmonic series, the terms ( \frac{1}{n} ) satisfy these conditions: they decrease steadily and approach zero as ( n ) increases. Consequently, the partial sums of the alternating harmonic series oscillate and converge to a specific limit, approximately 0.693147, which is the natural logarithm of 2 (( \ln 2 )).

Steps: Applying the Alternating Series Test To rigorously determine convergence, we apply the Alternating Series Test. This test requires two conditions for a series ( \sum (-1)^n a_n ) (where ( a_n > 0 )):

1. The sequence ( a_n ) must be monotonically decreasing: ( a_{n+1} \leq a_n ) for all ( n ) beyond some point.
2. The sequence ( a_n ) must approach zero: ( \lim_{n \to \infty} a_n = 0 ).

Step 1: Check the Monotonicity Condition Consider the sequence ( a_n = \frac{1}{n} ). Is ( \frac{1}{n+1} \leq \frac{1}{n} ) for all ( n )? Since ( n+1 > n ), it follows that ( \frac{1}{n+1} < \frac{1}{n} ). This holds true for all ( n \geq 1 ). Therefore, the sequence ( a_n = \frac{1}{n} ) is strictly decreasing.

Step 2: Check the Limit Condition Consider the limit of ( a_n = \frac{1}{n} ) as ( n ) approaches infinity: ( \lim_{n \to \infty} \frac{1}{n} = 0 ). This limit exists and equals zero.

Conclusion of the Test Since both conditions of the Alternating Series Test are satisfied, the alternating harmonic series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} ) converges.

Scientific Explanation: The Role of Partial Sums The convergence of the alternating harmonic series is elegantly demonstrated by examining its partial sums. Define the partial sum ( s_n ) as the sum of the first ( n ) terms:

• ( s_1 = 1 )
• ( s_2 = 1 - \frac{1}{2} = \frac{1}{2} )
• ( s_3 = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6} )
• ( s_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} )
• ( s_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{47}{60} )

A clear pattern emerges. The even-indexed partial sums (( s_2, s_4, s_6, \ldots )) form an increasing sequence: ( \frac{1}{2}, \frac{7}{12}, \frac{49}{60}, \ldots ). The odd-indexed partial sums (( s_1, s_3, s_5, \ldots )) form a decreasing sequence: ( 1, \frac{5}{6}, \frac{47}{60}, \ldots ). Crucially, every even-indexed partial sum is less than every odd-indexed partial sum. This is because adding the next term (( -\frac{1}{n} )) to an even sum (( s_{2k} )) gives the next odd sum (( s_{2k+1} )), and since ( \frac{1}{2k+1} > 0 ), it follows that ( s_{2k} > s_{2k+1} ).

The Boundedness and Convergence The sequence of even partial sums ( s_{2k} ) is bounded above (by ( s_1 = 1 )) and increasing, which implies it converges to some limit ( L ). The sequence of odd partial sums ( s_{2k+1} ) is bounded below (by ( s_2 = \frac{1}{2} )) and decreasing, implying it converges to some limit ( M ). Since each odd partial sum is obtained from the preceding even partial sum by subtracting the next positive term (( \frac{1}{2k+1} )), we have ( s_{2k+1} = s_{2k} - \frac{1}{2k+1} ). Taking

The convergence of the alternating harmonic seriesis elegantly demonstrated by examining its partial sums. Define the partial sum ( s_n ) as the sum of the first ( n ) terms:

• ( s_1 = 1 )
• ( s_2 = 1 - \frac{1}{2} = \frac{1}{2} )
• ( s_3 = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6} )
• ( s_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} )
• ( s_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{47}{60} )

A clear pattern emerges. The even-indexed partial sums (( s_2, s_4, s_6, \ldots )) form an increasing sequence: ( \frac{1}{2}, \frac{7}{12}, \frac{49}{60}, \ldots ). The odd-indexed partial sums (( s_1, s_3, s_5, \ldots )) form a decreasing sequence: ( 1, \frac{5}{6}, \frac{47}{60}, \ldots ). Crucially, every even-indexed partial sum is less than every odd-indexed partial sum. This is because adding the next term (( -\frac{1}{n} )) to an even sum (( s_{2k} )) gives the next odd sum (( s_{2k+1} )), and since ( \frac{1}{2k+1} > 0 ), it follows that ( s_{2k} > s_{2k+1} ).

The Boundedness and Convergence The sequence of even partial sums ( s_{2k} ) is bounded above (by ( s_1 = 1 )) and increasing, which implies it converges to some limit ( L ). The sequence of odd partial sums ( s_{2k+1} ) is bounded below (by ( s_2 = \frac{1}{2} )) and decreasing, implying it converges to some limit ( M ). Since each odd partial sum is obtained from the preceding even partial sum by subtracting the next positive term (( \frac{1}{2k+1} )), we have ( s_{2k+1} = s_{2k} - \frac{1}{2k+1} ). Taking the limit as ( k \to \infty ), the term ( \frac{1}{2k+1} \to 0 ). Therefore, ( L = M ), as the difference ( s_{2k+1} - s_{2k} \to 0 ). Thus, both subsequences converge to the same limit ( L ), confirming that the entire sequence of partial sums ( s_n ) converges to ( L ).

Scientific Explanation: The Role of Partial Sums The convergence of the alternating harmonic series is elegantly demonstrated by examining its partial sums. Define the partial sum ( s_n ) as the sum of the first ( n ) terms:

• ( s_1 = 1 )
• ( s_2 = 1 - \frac{1}{2} = \frac{1}{2} )
• ( s_3 = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6} )
• ( s_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} )
• ( s_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{47}{60} )

A clear pattern emerges. The even-indexed partial sums (( s_2, s_4, s_6, \ldots )) form an increasing sequence: ( \frac{1}{2}, \frac{7}{12}, \frac{49}{60}, \ldots ). The odd-indexed partial sums (( s_1, s_3, s_5, \ldots )) form a decreasing sequence: ( 1, \frac{5}{6}, \frac{47}{60}, \ldots ). Crucially, every even-indexed partial sum is less than every odd-indexed partial sum. This is because adding the next term (( -\frac{1}{n} )) to an even sum (( s_{2k} )) gives the next odd sum (( s_{2k+1} )),