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Oct 20 at 20:52 history edited LSpice CC BY-SA 4.0
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Oct 20 at 5:05 comment added Fedor Petrov Certainly: negative terms start with $k=n+3$, and they are way too small than the $k=2$ term, by the integral bound or whatever.
Oct 20 at 5:01 comment added user90533 Is this sum $\sum_{k=2}^{\infty}\left(\frac{n+2}{k^{n+3}}-\frac{1}{k^{n+2}}\right) $ positive for $n\in \mathbb{N}?$
Oct 20 at 2:51 history undeleted user90533
Oct 19 at 17:26 history deleted user90533 via Vote
Oct 19 at 11:38 comment added Fedor Petrov You may simply use the definition of $\zeta(s)=\sum1/n^s$, after cancellation the remaining positive terms are much larger in absolute values then negative
S Oct 19 at 9:06 review First questions
Oct 19 at 13:06
S Oct 19 at 9:06 history asked user90533 CC BY-SA 4.0