An inequality improvement on AMM 11145 I have asked the same question in math.stackexchange, I am reposting it here, looking for answers:
How to show that for $a_1,a_2,\cdots,a_n >0$ real numbers and for $n \ge 3$: 
$$\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le\left(2-\dfrac{7\ln{2}}{8\ln{n}}\right)\sum_{k=1}^{n}\dfrac{1}{a_{k}}$$
 A: (extended comment)
Choose some positive numbers $b_1,\dots,b_n$ and denote $a_i=ic_i$ for $i=1,\dots,n$. Use Cauchy-Bunyakovsky-Schwarz inequality $$\left(\sum_{i=1}^k b_ic_i\right)\left(\sum_{i=1}^k b_ic_i^{-1}\right)\geq \left(\sum_{i=1}^k b_i\right)^2$$ to estimate
$$
\sum_{k=1}^n \frac{k}{\sum_{i=1}^k b_ic_i}\leq \sum_{k=1}^n \sum_{i=1}^k \frac{k b_i c_i^{-1}}{(b_1+\dots+b_k)^2}=\sum_{i=1}^n c_i^{-1}b_i\sum_{k=i}^n\frac{k}{(b_1+\dots+b_k)^2}.
$$
Thus if some positive number $\alpha$ satisfies 
$$
b_i^2\sum_{k=i}^n\frac{k}{(b_1+\dots+b_k)^2}\leq \alpha,i=1,\dots,n,
$$
then we get desired inequality 
$$
\sum_{k=1}^n \frac{k}{\sum_{i=1}^k a_i}\leq \alpha \sum_{i=1}^n \frac1{a_i}.
$$
Choice of $b_i=i$ gives $\alpha=2$ or bit better, but there are rooms for improvement, since for large values $i$ (close to $n$) inequality becomes not very much sharp. 
Say, we may try $b_k=k+\lambda$ and optimize by $\lambda$. 
A: (This doesn't answer the actual question, but it was too long for a comment.Besides, it's related, and someone might even find it relevant.)
The following attempts $(1)$ to prove a non-discrete counterpart using integrals to the AMM 11145 inequality for an arbitrary real function $a(x)$ that is integrable and strictly positive on $[0, n]$:
$$
\int_0^n \frac{k}{\int_0^k a(x)\,dx}\,dk\leq 2\int_0^n \frac{1}{a(k)}\,dk
$$
and also $(2)$ prove that $2$ is the smallest constant for which it holds.
The notations are to emphasize the relationship to the discrete case, but $n, k, x, a(x)$ are all reals, of course. An equivalent integral inequality for $[0,\infty)$ was mentioned by @1015 here https://math.stackexchange.com/questions/599999/the-series-sum-limits-n-1-infty-frac-n-frac1a-1-frac1a-2-dotsb-fra/600943#600943 but was not elaborated or further pursued at the time.
$(1)$ The Cauchy-Schwartz inequality for real-valued functions
$$
\left\lvert{\int_0^k f(x)\,g(x)\,dx}\right\rvert^2 \le \int_0^k f(x)^2 \,dx \int_0^k g(x)^2 \,dx 
$$
written for $f(x) = x / {\sqrt{(a(x))}}$ and $g(x) = \sqrt{(a(x))}$ gives
$$
\left\lvert{\int_0^k{x\,dx}}\right\rvert^2 \le \int_0^k \frac{x^2}{a(x)}\,dx\;\int_0^k a(x)\,dx
$$
then replacing the left hand side with the obvious $k^4 / 4$ and rearranging
$$
\frac{k}{\int_0^k a(x)\,dx} \le \frac{4}{k^3}\;\int_0^k\frac{x^2}{a(x)}\,dx
$$
Integrating in $k$ over $[0, n]$ then applying Fubini's theorem and simplifying gives
$$
\begin{align}
\int_0^n \frac {k} {\int_0^k a(x)\,dx}\,dk & \le \int_0^n \int_0^k\frac{4}{k^3}\; \frac{x^2}{a(x)}\,dk\,dx \\
& = \int_0^n \int_x^n\frac{4\,x^2}{a(x)}\; \frac{1}{k^3}\,dx\,dk \\
& = \int_0^n \frac{4\,x^2}{a(x)}\; (\frac{1}{2\,x^2} - \frac{1}{2\,n^2})\,dx \\
& = 2\;\int_0^n \frac{1}{a(x)}\; (1 - \frac{x^2}{n^2})\,dx \\
& \le 2\;\int_0^n \frac{1}{a(x)}\,dx = 2\;\int_0^n \frac{1}{a(k)}\,dk
\end{align}
$$
which proves the claimed inequality.
$(2)$ To prove that $C = 2$ is the best constant, consider the function $a(x) = x + \epsilon$. After calculating the elementary integrals, the inequality reduces to:
$$
2\,(\,ln(n + 2\,\epsilon) - ln(2\,\epsilon)\,) \le 2\,(\,ln(n + \epsilon) - ln(\epsilon))
$$
The ratio of the two sides approaches $1$ as $\epsilon \to 0$ as can be easily verified using l'Hopital rule for example:
$$
\begin{align}
\lim_{\epsilon\to 0}{ln(n + 2\,\epsilon) - ln(2\,\epsilon) \over ln(n + \epsilon) - ln(\epsilon)} = {\frac{2}{n+2\,\epsilon} - \frac{2}{2\,\epsilon} \over \frac{1}{n+\epsilon} - \frac{1}{\epsilon}} = {\frac{-\,n}{n + 2\,\epsilon} \over \frac{-\,n}{n+\epsilon}} = 1
\end{align}
$$
...which indicates that any constant $C' < 2$ would fail to satisfy the inequality for small enough $\epsilon$.
To sum it up, the result itself is not surprising. The proofs for the discrete case inequality usually showed that $2$ is the best constant for arbitrarily large series as well. The formulation using integrals just allows for the equivalent of such infinite series to occur within a finite $[0,n]$ interval. It might be also worth noting that the linear function used in the second step reminisces of the proofs by harmonic numbers in the discrete case (that $2$ is optimal for infinite series).
A: I came up with something years ago which is similar to Fedor Petrov's (or other users'). 
Problem: Let $a_i > 0; \ i = 1, 2, \cdots, n$ ($n\ge 2$). Let $C(n) = 2 - \frac{7\ln 2}{8\ln n}$. Prove that
$$\sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k}
\le C(n)\sum_{k=1}^n \frac{1}{a_k}.$$
Introducing the coefficients (to be determined) $C_1, C_2, \cdots, C_n > 0$, by the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\frac{k}{a_1 + a_2 + \cdots + a_k} \le \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}$$
and
\begin{align}
\sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k} 
&\le \sum_{k=1}^n \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}\\
&=  \sum_{k=1}^n \Big[C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\Big]\frac{1}{a_k}.
\end{align}
Equality occurs if and only if $C_k = a_k, \ k=1, 2, \cdots, n$.
The problem becomes: Can we choose $C_k > 0,\ k=1, 2, \cdots, n$ such that 
$$\sup_{k=1, 2, \cdots, n} C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} \le C(n), \ \forall n\ge 2 ?$$
It is easy if $C(n)=2$. For a simple proof, I chose $$C_k = \sqrt{k(k+1)(k+2)(k+3)} - \sqrt{(k-1)k(k+1)(k+2)}, \ k = 1, 2, \cdots, n.$$
We have $C_1 + C_2 + \cdots + C_m = \sqrt{m(m+1)(m+2)(m+3)}$ and
\begin{align}
&C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\\
= \ &k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big)\\
&\quad \cdot\Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)} - \frac{1}{2(n+2)(n+3)}\Big)\\
\le \ & k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big)
\Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)}\Big)\\
\le \ & 2.
\end{align}
Fedor Petrov chose $C_k = k,\ \forall k$. We have
$$C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} = 
k^2\Big(-\frac{4}{n+1}+4\Psi(1, n+2)+\frac{4}{k}-4\Psi(1, 1+k)\Big).$$
Denote the RHS as $f(k, n)$. 
We have the asymptotic expansion $f(\sqrt{n}, n) \sim 2 - \frac{2}{3\sqrt{n}} - \frac{2}{n} + \cdots (n\to \infty)$.
Thus, it at most gives $C(n) \ge 2 - \frac{A}{\sqrt{n}}$ where $A$ is a constant.
Fedor Petrov also suggested $C_k = k + \lambda$ (optimize $\lambda$). If $\lambda$ is a constant, similarly,
we have $f(\sqrt{n},n, \lambda) \sim 2 - \frac{4\lambda + 2}{3\sqrt{n}} - \cdots (n \to \infty)$.
It is not enough. 
