DylsexicChciken Posted May 2, 2015 Share Posted May 2, 2015 (edited) The alternative series test says if the absolute value of the terms are not increasing and the limit of the absolute value terms goes to 0, then the alternating series converges. What happens if the converse is satisfied, when the absolute value terms of an alternating series are increasing or the limit of the absolute value terms does not go to zero? If either test fails, can we say the alternating series is divergent? Edited May 2, 2015 by DylsexicChciken Link to comment Share on other sites More sharing options...

mathematic Posted May 2, 2015 Share Posted May 2, 2015 The alternative series test says if the absolute value of the terms are not increasing and the limit of the absolute value terms goes to 0, then the alternating series converges. What happens if the converse is satisfied, when the absolute value terms of an alternating series are increasing or the limit of the absolute value terms does not go to zero? If either test fails, can we say the alternating series is divergent? It may not be divergent, but it won't be convergent. The partial sums could oscillate. Example: 1-1+1-1+1-1........ Link to comment Share on other sites More sharing options...

DylsexicChciken Posted May 3, 2015 Author Share Posted May 3, 2015 It may not be divergent, but it won't be convergent. The partial sums could oscillate. Example: 1-1+1-1+1-1........ So basically the converse is true in that it won't converge, thanks. Link to comment Share on other sites More sharing options...

John Posted May 3, 2015 Share Posted May 3, 2015 What you're asking about is the inverse, not the converse, and the inverse is not necessarily true. For instance, consider [math]\sum_{n = 1}^{\infty} (-1)^n a_n[/math] where [math]a_n = \begin{cases} \frac{1}{n^2} & n \textnormal{ is odd} \\ \frac{1}{n^3} & n \textnormal{ is even}. \end{cases}[/math] Then our series is [math]\sum_{n = 1}^{\infty} (-1)^n a_n = -1 + \frac{1}{8} - \frac{1}{9} + \frac{1}{64} - \frac{1}{25} + \frac{1}{216} - \frac{1}{49} + \cdots[/math] which converges even though [math]a_n[/math] is not monotonically decreasing. The converse of the alternating series test would be that if the series is convergent, then the absolute values of the terms are monotonically decreasing and the limit is zero. Since the converse is the contrapositive of the inverse, we see that the converse is not necessarily true either. 2 Link to comment Share on other sites More sharing options...

DylsexicChciken Posted May 21, 2015 Author Share Posted May 21, 2015 (edited) What you're asking about is the inverse, not the converse, and the inverse is not necessarily true. For instance, consider [math]\sum_{n = 1}^{\infty} (-1)^n a_n[/math] where [math]a_n = \begin{cases} \frac{1}{n^2} & n \textnormal{ is odd} \\ \frac{1}{n^3} & n \textnormal{ is even}. \end{cases}[/math] Then our series is [math]\sum_{n = 1}^{\infty} (-1)^n a_n = -1 + \frac{1}{8} - \frac{1}{9} + \frac{1}{64} - \frac{1}{25} + \frac{1}{216} - \frac{1}{49} + \cdots[/math] which converges even though [math]a_n[/math] is not monotonically decreasing. The converse of the alternating series test would be that if the series is convergent, then the absolute values of the terms are monotonically decreasing and the limit is zero. Since the converse is the contrapositive of the inverse, we see that the converse is not necessarily true either. Would the inverse be true if the sequence is not a piece-wise function? It seems two textbooks I have read have been implying that the inverse is valid if the conditions for alternating test for convergence is not satisfied, albeit the textbooks does not deal with piece-wise series. Edited May 21, 2015 by DylsexicChciken Link to comment Share on other sites More sharing options...

John Posted May 21, 2015 Share Posted May 21, 2015 Hm. I think[math]a_n = \frac{\sin{n} + 1}{n}[/math] works. WolframAlpha seems to think it does, anyway. Link to comment Share on other sites More sharing options...

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