By non-trivial, I mean not only non-zero but also a neat looking one, so something you get by normalizing well known series wouldn't work, example : $\sum_{n=0}^{\infty}\left( \frac{1}{en!} - \frac{1}{2^{n+1}}\right)$.
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- $\begingroup$ Do you mean whose limits are equal to $0$? $\endgroup$J. W. Tanner– J. W. Tanner2020-09-07 15:54:29 +00:00Commented Sep 7, 2020 at 15:54
- $\begingroup$ @J.W.Tanner sure, Infinite series. $\endgroup$the_firehawk– the_firehawk2020-09-07 15:55:09 +00:00Commented Sep 7, 2020 at 15:55
- 1$\begingroup$ It could be useful to give a little proof or reference on answers if possible $\endgroup$Dabed– Dabed2020-09-07 22:46:46 +00:00Commented Sep 7, 2020 at 22:46
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3 Answers
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1 $$\sum_{k=0}^\infty(-1)^k\frac{\pi^{2k+1}}{(2k+1)!}.$$
- 1$\begingroup$ Is this $\sin\pi$? $\endgroup$Vincenzo Tibullo– Vincenzo Tibullo2020-09-07 15:57:41 +00:00Commented Sep 7, 2020 at 15:57
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$$ \sum_{n=1}^\infty \sin \left(\frac{\pi (2n+1)}{(n+1)n}\right) \sin\left(\frac{\pi}{(n+1)n}\right) $$
$$ \sum_{n=1}^\infty \frac{\left(\cos(\pi n/8) + \cos(7 \pi n/8) - 2 \cos(3 \pi n/8)\right)}{n^2} $$
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$$ \sum_{k=0}^{\infty}\binom{1/2}{k}(-1)^k=1-\frac{1}{2}-\frac{1}{8}-\frac{1}{16}-\frac{5}{128}-\frac{7}{256}-\dots=0 $$ (see Binomial series)
