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

Question

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 ?

Answer 1

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 $\frac{1}{p^{\alpha+1}}$.