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Does there exist a series conditionally convergent to 0 with sequence of partial sums {sn} such that sn is positive for every n? I considered 1-1+1/2 -1/2 +1/3 -1/3 + ....This series converges conditionally to zero but its sequence of partial sums has all odd terms positive and all even terms zero. The series (1-log2) -1/2 +1/3 -1/4 +.....which converges to zero conditionally has sequence of partial sums with odd terms positive and even terms negative. The rearrangement of 1-1/2 +1/3 -1/4 .....with one positive term followed by 4 negative terms which conditionally converges to zero is also having its sequence of partial sums with four positive terms followed by a negative 5th term and so on alternately. These examples make me believe that there does not exist a conditionally convergent series with all its sequence of partial sums positive. How to prove this or how to get a counter example? Please help!

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Yes, there is:$$1-\frac34+1-1+\frac14-\frac38+\frac12-\frac12+\frac18-\frac3{16}+\frac13-\frac13+\frac1{16}-\cdots,$$whose sum is $0$ and whose partial sums are$$1,\frac14,1+\frac14,\frac14,\frac12,\frac18,\frac12+\frac18,\frac18,\frac14,\frac1{16},\frac13+\frac1{16},\frac1{16},\ldots$$

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  • $\begingroup$ Thank you. The series conditionally converges to zero and has a positive sequence of partial sums. $\endgroup$ – Lawrence Mano Apr 5 '20 at 23:23
  • $\begingroup$ I forgot to check that the series given is not absolutely convergent. It looks like the series is absolutely convergent because the series of positive terms as well as the series of negative terms both converge yielding absolute convergence. I am afraid this is not the correct example. Please correct me if I’m wrong. $\endgroup$ – Lawrence Mano Apr 6 '20 at 0:22
  • $\begingroup$ You are right. My bad. I've edited my answer, adding terms like $\frac1n-\frac1n$ in the middle of my series, This is a universal method: it turns any convergent series into a conditionally convergent one. $\endgroup$ – José Carlos Santos Apr 6 '20 at 7:08
  • $\begingroup$ Now it is perfectly clear to me. Thank you so much. $\endgroup$ – Lawrence Mano Apr 7 '20 at 2:44
  • $\begingroup$ I'm glad I could help. Besides, it was funny to solve this problem. $\endgroup$ – José Carlos Santos Apr 7 '20 at 6:50

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