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For $\alpha >0$, I have to show that $$\sum_{n=1}^\infty \frac{(-1)^{\frac{n(n+1)}{2}}}{n^\alpha }$$ is convergent. In my book, if $u_n=\frac{(-1)^{\frac{n(n+1)}{2}}}{n^\alpha }$, they set

$$v_p=u_{4p}+u_{4p+1}+u_{4p+2}+u_{4p+3},$$ and they proved that

$$v_p=\frac{1}{(4p)^\alpha }\left(\frac{-\alpha }{p}+\mathcal O\left(\frac{1}{p}\right)\right)$$ and says at the end (see the picture below)

$$\frac{1}{(4p)^\alpha }\left(\frac{-\alpha }{p}+\mathcal O\left(\frac{1}{p}\right)\right)\sim_{p\to \infty }\frac{-\alpha }{4^\alpha p^{\alpha +1}}.$$

But this is not correct, is it ?

enter image description here

Peter
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  • I think you're confusing little o and big O. – Bernard Jul 30 '18 at 20:54
  • @Bernard: I'm not. As you can see it's written in my book. But they probably wanted to write a $o$ instead of a $O$, no ? – Peter Jul 30 '18 at 20:57
  • I guess so too. – Bernard Jul 30 '18 at 20:59
  • @Peter Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 06 '18 at 20:19

1 Answers1

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It should be a little-o notation, then just multiply to obtain

$$\frac{1}{(4p)^\alpha }\left(\frac{-\alpha }{p}+\mathcal o\left(\frac{1}{p}\right)\right)=\frac{-\alpha }{4^\alpha p^{\alpha +1}}+\mathcal o\left(\frac{1}{p^{\alpha+1}}\right)$$

and $\mathcal o\left(\frac{1}{p^{\alpha+1}}\right)\to 0$ more rapidly then $\frac{1}{p^{\alpha+1}}$.

user
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