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May 2, 2022 at 6:38 history edited Random Variable CC BY-SA 4.0
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May 2, 2022 at 6:33 history edited Random Variable CC BY-SA 4.0
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Jun 28, 2014 at 16:50 history edited Random Variable CC BY-SA 3.0
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May 11, 2014 at 16:13 history bounty awarded CommunityBot
May 9, 2014 at 16:44 comment added Random Variable @robjohn Thanks.
May 9, 2014 at 16:38 comment added robjohn @user91500: you can pull out the $k=0$ term and handle it separately. In both sums that is $$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$ Then the double sums for $n\ge0$ and $k\ge1$ converge absolutely for $|z|\lt\frac\pi2$.
May 9, 2014 at 15:45 comment added Random Variable @user91500 The justification might be nontrivial. That's why I didn't post that second evaluation until Peter asked about a different approach. Sufficient, but not necessary, conditions is if $$\sum_{n=0}^{\infty} \sum_{k=0}^{\infty}|\frac{2^{2k} z^{2k}}{\pi^{2k}} \frac{(-1)^{n}}{(2n+1)^{2k+1}}| $$ or $$\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} |\frac{2^{2k} z^{2k}}{\pi^{2k}} \frac{(-1)^{n}}{(2n+1)^{2k+1}}|$$converges. The second iterated sum does converge if $k$ starts from $k=1$. That might show that the sum is well-behaved enough to switch the order of summation. Let me ask around.
May 9, 2014 at 14:11 comment added user91500 $\sum_{n=0}^\infty\sum_{k=0}^\infty|(-1)^n\frac{z^{2k}}{{(2n+1)^{2k+1}}}|$ is divergent, How did you switch the order of summation here?
May 7, 2014 at 4:19 comment added Random Variable @Peter I added a second proof to my answer.
May 7, 2014 at 4:16 history edited Random Variable CC BY-SA 3.0
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Apr 30, 2014 at 16:44 vote accept CommunityBot
Apr 21, 2014 at 22:21 history edited Random Variable CC BY-SA 3.0
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Apr 21, 2014 at 17:33 history edited Random Variable CC BY-SA 3.0
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Apr 21, 2014 at 11:32 history edited Random Variable CC BY-SA 3.0
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Apr 21, 2014 at 11:24 history edited Random Variable CC BY-SA 3.0
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Apr 21, 2014 at 11:15 history answered Random Variable CC BY-SA 3.0