Let $\{d_t\}_{t \geq 1}$ be a sequence of real numbers with each $d_t \in [0, 1]$. Suppose that we know that: $$ \lim_{T \to \infty} \frac{1}{T} \sum_{t=1}^{T} d_t = 0 \:. \quad\quad (1) $$
Fix a constant $\alpha \in (0, 1)$. Is it necessarily true that there must exist an $\beta \in (0, 1)$ such that: $$ \limsup_{T \to \infty} \max_{k=\lceil \alpha T \rceil, ..., T} \frac{1}{(T-k+1)^\beta} \sum_{t=k}^{T} d_t < \infty \:. \quad\quad (2) $$
For example, consider the sequence with $d_t = 1/\sqrt{t}$. Then: $$ \sum_{t=1}^{T} d_t = \sum_{t=1}^{T} \frac{1}{\sqrt{t}} \leq 2 \sqrt{T} \:, $$ and so condition (1) holds. On the other hand: $$ \sum_{t=k}^{T} d_t = \sum_{t=k}^{T} \frac{1}{\sqrt{t}} \leq 2 (\sqrt{T} - \sqrt{k-1}) \:. $$ By the concavity of $x \mapsto \sqrt{x}$, we can upper bound for $k \geq 2$: $$ \sqrt{T} - \sqrt{k-1} \leq \frac{1}{2\sqrt{k-1}} (T - k + 1) \:. $$ Hence when $k \geq \alpha T$: $$ \frac{1}{\sqrt{T-k+1}} \sum_{t=k}^{T} d_t \leq \frac{\sqrt{T-k+1}}{\sqrt{k-1}} = \sqrt{\frac{T}{k-1} - 1} \leq \sqrt{\frac{1}{\alpha} \frac{T}{T-1/\alpha} - 1} \:. $$ It is clear that this bound is finite as $T \to \infty$. Hence (2) holds with $\beta=1/2$.
My question is if this type of behavior is general or is there a counter-example when this does not occur?
Here is the counter example.
We generate a sequence $\{d_t\}$ according to the following procedure. Initially set all $d_t$'s equal to zero. Now for $k=1, 2, ...$, we set $d_t = 1$ for all $t \in [2^k - (k-1), 2^k]$.
First, (1) holds since $\sum_{t=k}^{T} d_t$ is bounded by roughly $\log_2^2(T)$, since at position $T$ there are at most $\lceil{\log_2(T)\rceil}$ subsequences of consecutive 1's in $\{d_t\}_{t=1}^{T}$, each of length at most $\log_2(T)$.
Now (2) clearly does not hold, since when $T=2^i$ for some $i$, then for $k=T - \log_2(T) + 1$ the tail sum $\sum_{t=k}^{T} d_t = T - k$.