Asymptotic behavior of $\sum_{t=1}^T \frac{s_t}{t}$, when $\sum_{i=1}^T s_i = A$

Question

consider the harmonic series: $$ h_T = \sum_{t=1}^T \frac{1}{t} $$ We know that $h_T \in O(\log T)$. Now consider a modified version of the harmonic series: $$ g_T = \sum_{t=1}^T \frac{s_t}{t} $$ where $\sum_{i=1}^T s_i = A$. Is there an easy way to bound upper bound $g_T$ as a function of $A$ and $T$?

Answer 1

What might be useful is Chebychev's sum inequality: https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality

If the $s_i$ are decreasing, since $\frac1{t}$ is decreasing, this says that $\frac1{T}\sum_{t=1}^T \frac{s_t}{t} \ge \left(\frac1{T}\sum_{t=1}^T s_t\right)\left( \frac1{T}\sum_{t=1}^T \frac1{t}\right) = \frac{A}{T}\frac{h_T}{T} $ or $g_T \ge \frac{A h_T}{T} $.

If the $s_i$ are increasing, this says that $\frac1{T}\sum_{t=1}^T \frac{s_t}{t} \le \frac1{T}\sum_{t=1}^T s_t \frac1{T}\sum_{t=1}^T \frac1{t} = \frac{A}{T}\frac{h_T}{T} $ or $g_T \le \frac{A h_T}{T} $.