There is an exercise in the book on page $106$ that given a multiplicative function $h$ satisfying the conditions ($p$ denotes a prime)
$$|h(p)|\le 1 \text{ if } (p|N), |h(p)|\le p^{-\delta} \text{ if } (p\nmid N), |h(p^\nu)|\ll 1 \text{ if } (\nu\ge 2),$$
where $\delta$ is a positive constant. Show that, for $0\le \alpha\le 1/\log\log N$, we have $$\sum_{d\ge 1}|h(d)|d^{\alpha-1}\ll \log\log N.$$ I wonder that shouldn't the right part of the problem be just $1$ instead of $\log\log N$? Why is it needed to have a condition that $\alpha\le 1/\log\log N$?
Here are what I'm trying to show that the right part is $1$ but I'm confused (both the usefulness of the above problem to the other problem that the book tried to guide and it is much comprehensible if it is $1$ as it click with the help of the following problem, and the proof of it.)
\begin{align*} \sum_{d\le P}\dfrac{|h(d)|}{d^{1-\alpha}}&\le \prod_{p\le P}\left(1+\dfrac{|h(p)|}{p^{1-\alpha}}+\dfrac{|h(p^2)|}{p^{2(1-\alpha)}}+\cdots\right)\\ &\le \prod_{\substack{p\le P\\ p\nmid N}}\left(1+\dfrac{1}{p^\delta(p^{1-\alpha})}+\left\{\dfrac{C}{p^{2(1-\alpha)}}+\dfrac{C}{p^{3(1-\alpha)}}+\cdots\right\}\right)\prod_{\substack{p\le P\\ p|N}}\left(1+\dfrac{1}{p^{1-\alpha}}+\dfrac{C'}{p^{2(1-\alpha)}}+\cdots\right), \end{align*} as we have the theorem that $\prod (1+a_n)<\infty\Longleftrightarrow \sum a_n<\infty$, we have that the above converges as for large $N$, $\alpha\le 1/\log\log N$ will eventually $<\delta$ and thus $$C\sum_{p}\sum_{k\ge 2}\dfrac{1}{p^{k(1-\alpha)}}+\sum_{p}\dfrac{1}{p^{1+\delta-\alpha}}<\infty.$$ Therefore, the first product converges and the second has only finite terms so $$\sum_{d\ge 1}\dfrac{|h(d)|}{d^{1-\alpha}}\ll 1.$$