For any positive integer $n$, what is the value of $t^*$ that maximises the following expression? For any positive integer n, what is the value of t* that maximises the following expression? 
$$\displaystyle \sum_{j=1}^{n-t^*}\left(\frac{t^*-j+2}{t^*+j}\right)$$
where $t^*$ is some integer in the set $\{0,1,2...,n-1\}$.
Clearly $t^*$ = $f(n)$ but I am unable to find what this function is? E.g. $t^*=\frac{3}{5}(n-1)$ rounded to the nearest whole number is a decent estimate but not exact. Perhaps there is no closed-form solution. Thanks for your help!
 A: Reindex with $j\mapsto j-t^*$ and you have $$\sum_{j=1+t^*}^n\frac{2t^*-j+2}{j}$$ or just  $$\sum_{j=1+t^*}^n\left(\frac{2t^*+2}{j}-1\right)$$ which is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)$$
The difference between this expression and the same expression with $t^*$ replaced by $t^*-1$ is $$(2t^*+2)\sum_{j=1+t^*}^n\frac{1}{j}-(n-t^*-1)-2t^*\sum_{j=t^*}^n\frac{1}{j}+(n-t^*)$$ which simplifies to $$2\sum_{j=1+t^*}^n\frac{1}{j}-1$$ The critical value for $t^*$ will happen when this difference is $0$. Either the floor or the ceiling of that number is the answer you are seeking. Representing $H(m)$ as the $m$th partial Harmonic summ, we are trying to solve $$2H(n)-2H(t^*)=1$$ or rather $$H(t^*)=H(n)-\frac{1}{2}$$ If you can find an inverse function for $H$, you have your answer. For large $n$, $H(n)\approx \log(n)+\gamma$, where $\gamma$ is the Euler-Mascheroni constant, so if we take that and run with it, $$t^*\approx\exp\left(\log(n)-\frac{1}{2}\right)=\frac{1}{\exp(1/2)}n\approx0.6065n$$ agreeing with your empirical estimate of $3/5$. This is more accurate the larger the value of $n$.
A: In the edit of my answer to this question, I proposed for the inverse of the harmonic number
$$n\sim -\frac{1}{2}+\sqrt{\frac{1}{6 W(k)}-\frac{1}{4}}\quad \text{where} \qquad k=\frac{1}{2}e^{2(\gamma-H_n)}$$
So, after @Alex Jordan's answer,
$$H(t^*)=H(n)-\frac{1}{2} \implies t^*=-\frac{1}{2}+\frac{1}{6} \sqrt{\frac{6-9 W\left(\frac{1}{6} e^{1-2 \psi
   (n+1)}\right)}{W\left(\frac{1}{6} e^{1-2 \psi
   (n+1)}\right)}}$$ Expanding as a series for large $n$, this gives
$$t^*=\frac{n}{\sqrt{e}}\Bigg[1+\frac{1-\sqrt{e}}{2 n}+\frac{1-e}{24 n^2}+\frac{e-1}{48
   n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$
For $n=10$, this would give $$t^*=\frac{50419-2400 \sqrt{e}-19 e}{4800 \sqrt{e}}=5.864447$$ while the exact solution is
$5.864453$
