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$$\sum_{k=0}^{\infty} 1^k-1^{2k}.$$ On the one hand partial sums of this series are equal to 0, so our infinite series must converge to 0. On the other hand $$\sum_{k=0}^{\infty} 1^k-1^{2k}= \sum_{k=0}^{\infty} 1^k-\sum_{k=0}^{\infty} 1^{2k}=\sum_{k=0}^{\infty} 1^{2k+1}$$ Obviously this series is not converge to 0. And i cant understand what rules i'm breaking?

Matti P.
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Arial Pilisov
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1 Answers1

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$\sum_{k=0}^{\infty} 1^k-1^{2k}= \sum_{k=0}^{\infty} 1^k-\sum_{k=0}^{\infty} 1^{2k}\neq\sum_{k=0}^{\infty} 1^{2k+1}$

I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?

KnowsNothing
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  • Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1. – Arial Pilisov Nov 21 '18 at 08:30
  • Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish – KnowsNothing Nov 21 '18 at 08:37
  • I suppose that Yves Daoust's answer to this question https://math.stackexchange.com/questions/657241/infinite-series-1-frac12-frac23-frac14-frac15-frac26-cdots?rq=1 is a good example of this kind of stuff. – KnowsNothing Nov 21 '18 at 08:41
  • Thank you very much. I got it. – Arial Pilisov Nov 21 '18 at 18:57