Let $\{a_k\}$ be a non-increasing sequence tending to zero. Suppose that $\sum_{k} a_k^\theta < \infty$ for some $\theta < 2$. Show that $\sum_{n=1}^\infty (\sum_{k=1}^n a_k^{-2} )^{-1/2} < \infty$.
The text I am reading claims that this is trivial and omit the proof. But, obviously I am too laggard to see the trick. Can anyone give some hint? Thanks for your help!
Using that $(a_k)_{k \in \mathbb{N}}$ is decreasing, we get that $$\sum_{k=1}^n a_k^{-2} \geq \sum_{k=\lfloor n/2 \rfloor}^n a_k^{-2} \geq \frac{n}{2} (a_{\lfloor n/2 \rfloor})^{-2}.$$ Thus $$\sum_{n=1}^\infty \Big( \sum_{k=1}^n a_k^{-2} \Big)^{-1/2} \ll \sum_{n=1}^\infty n^{-1/2} a_n.$$ Now, apply the Hölder-inequality with $p=\theta$ and $q= \theta/(\theta-1)>2$. Note that $\sum_{n=1}^\infty n^{-q/2}$ is convergent, because $q>2$.
Edit: One case is missing. Here, I have supposed that $\theta >1$, otherwise we can just use $n^{-1/2} a_n \leq a_n^\theta$.