How to prove/disprove $ \sum_{i=1}^{n} \frac{a_i}{ \sum_{j=1}^{i} a_j } \approx \log \sum_{i=1}^{n} a_i, \quad a_i \in \mathbb{N}^+ $?

Question

Remember $\sum_{i=1}^{n} 1/i$ is asymptotic to $\log n$. Is it possible to generalize it to the following?: $$ \sum_{i=1}^{n} \frac{a_i}{ \sum_{j=1}^{i} a_j } \approx \log \sum_{i=1}^{n} a_i, \quad a_i \in \mathbb{N}^+ $$

Answer 1

We have inequality $$\frac{1}{x+1} \leq \ln \left(1 +\frac{1}{x} \right) \leq \frac{1}{x} $$ Denote $s_j =\sum_{k=1}^j a_k $ then $$\sum_{t=2}^n \frac{a_t }{s_{t-1}} =\sum_{t=2}^n \frac{1 }{\frac{s_{t-1}}{a_t}}\geq \sum_{t=2}^n \ln \left( 1 +\frac{1 }{\frac{s_{t-1}}{a_t}}\right)=\sum_{t=2}^n \ln\left(\frac{s_{t}}{s_{t-1}}\right)=\ln\left(\frac{s_{n+1}}{s_1}\right)$$ on the other hand $$\sum_{t=1}^n \frac{a_t }{s_{t}} =\sum_{t=1}^n \frac{1 }{\frac{s_{t-1}}{a_t}+1}\leq \sum_{t=1}^n \ln \left( 1 +\frac{1 }{\frac{s_{t-1}}{a_t}}\right)=\sum_{t=1}^n \ln\left(\frac{s_{t}}{s_{t-1}}\right)=\ln\left(\frac{s_{n+1}}{s_1}\right)$$