On an asymptotic improvement of AMM problem 11145 (April 2005)

Question

Motivation
Motivated by this question, I tried improve the inequality $$\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le2\sum_{k=1}^{n}\dfrac{1}{a_{k}}$$ asymptotically. In other words, with support of some numerical evidence, I want to find the value of the following limit.

Question
Numerical experiment indicates that $$\lim_{n\to \infty}\ln n\left( 2-\sup_{a_k>0 (k=1\ldots n)}\frac{\sum_{k=1}^n{\frac{k}{a_1+a_2+\cdots +a_k}}}{\sum_{k=1}^n{1/a_k}} \right) $$ exists. The limit seems to be approximately $1.5$. How can we prove it and find its value?

Some Trivial Results
From this answer we can see that this limit is of form $\infty\cdot0$.
One can apply $\frac{\partial}{\partial a_k}$ to the formula in the $\sup$ and get a simultaneous equation, which is extremely complex and hence almost unsolvable with unknown $n$.

Answer 1

This is not an answer, but an experiment for moderately large $n$, making the limit of $\approx 1.5$ unlikely. $$S_n=\sup\left\{\sum_{k=1}^{n}\frac{k}{\sum_{j=1}^{k}a_j} : a_1,\ldots,a_n>0,\ \sum_{k=1}^{n}\frac{1}{a_k}=1\right\}$$ The method of Lagrange works here. Writing $$\mathcal{L}(a_1,\ldots,a_n,\lambda)=\sum_{k=1}^{n}\frac{k}{\sum_{j=1}^{k}a_j}-\lambda\sum_{k=1}^{n}\frac{1}{a_k},\\\frac{\partial\mathcal{L}}{\partial a_m}=\frac{\lambda}{a_m^2}-\sum_{k=m}^{n}\frac{k}{\left(\sum_{j=1}^{k}a_j\right)^2}\quad(1\leqslant m\leqslant n),$$ we obtain the following system for $(a_1,\ldots,a_n,\lambda)$: $$\sum_{k=1}^n\frac{1}{a_k}=1;\quad\frac{k}{\left(\sum_{j=1}^{k}a_j\right)^2}=\frac{\lambda}{a_k^2}-\frac{\lambda}{a_{k+1}^2}\quad(1\leqslant k\leqslant n)$$ where we set $a_{n+1}:=\infty$ so that the very last term vanishes when $k=n$.

For fixed $\lambda$, the ratios $a_{k+1}/a_1$ can be computed from the second equation taken at $1\leqslant k\leqslant n-1$; then, the first equation determines $a_1$, and the last equation at $k=n$, finally, can be viewed as a single equation for $\lambda$. Observe also that if $(a_1,\ldots,a_n,\lambda)$ is the solution, then summation by parts yields simply $$S_n=\sum_{k=1}^{n}\left(\frac{\lambda}{a_k^2}-\frac{\lambda}{a_{k+1}^2}\right)\sum_{j=1}^{k}a_j=\lambda\sum_{k=1}^{n}\frac{1}{a_k}=\lambda.$$

This handles $\color{red}{n\sim 10^6}$ easily. For powers of $2$, running

foo(n,t)={my(b=1,r=0,s=0);for(k=1,n,if(b<0,return(-1));my(a=sqrt(b));r+=a;s+=1/a;b-=k/t/s^2);return(b)};
for(n=1,20,printf("%2u... ",n);print(n*log(2)*(2-solve(t=1.1,2.0,foo(2^n,t)))))

in PARI/GP (under different precision settings for sure), I get

 1... 0.60650378298995214574[+++]
 2... 1.02433225912743257353[+++]
 3... 1.27241880529843009127[+++]
 4... 1.39985861561679724754[+++]
 5... 1.45256362011495626110[+++]
 6... 1.46164482213152111129[+++]
 7... 1.44578727489838766722[+++]
 8... 1.41589954712815847210[+++]
 9... 1.37839977433573904591[+++]
10... 1.33712471597920952819[+++]
11... 1.29439637045726263776[+++]
12... 1.25162537387072306196[+++]
13... 1.20966180786244720779[+++]
14... 1.16900506057647903574[+++]
15... 1.12993258591926390364[+++]
16... 1.09258058237638236418[+++]
17... 1.05699540545119896406[+++]
18... 1.02316675997623301901[+++]
19... 0.99104933243682303143[+++]
20... 0.96057697237757525984[+++]

Thus, the expression $L_n=(2-S_n)\ln n$ under the limit becomes $<1$. In fact $$L_{430605}=1.00000009015459070849\!+\\L_{430606}=0.99999998376961512393\!-$$