Suppose $x_t$ is a nonnegative sequence satisfying $$ \sum_{t=1}^{+\infty} x_t^2 < \infty.$$ I am trying to get a precise estimate for how fast $\sum_{t=1}^T x_t$ can grow as a function of $T$. Application of Cauchy-Schwarz gives that $$\sum_{t=1}^T x_t \leq \sqrt{T} \sqrt{\sum_{t=1}^{+\infty} x_t^2},$$ so $O(\sqrt{T})$ is one upper bound. My question is whether in fact $$ \lim_{T \rightarrow +\infty} \frac{1}{\sqrt{T}} \sum_{t=1}^T x_t = 0.$$
Here is why one might hope that such a thing is true. First, Cauchy-Schwarz is tight when the two vectors are multiples of each other, and since $x_t \rightarrow 0$, the vector $(x_1, \ldots, x_T)$ is very far from being a multiple of $(1,...,1)$. Second, if we try to come up with a tight example, the natural guess might be $x_t = 1/(\sqrt{t} \log^c(t))$ for some $c>0$, since its square is close to being the slowest decaying summable sequence. But in that case $\sum_{t=1}^T x_t = O(\sqrt{T}/\log(T))$, and the limit is indeed zero.