Estimate, or give an exact formula for $g(n) =\sum_{j=0}^{n} \dfrac{1}{(j+1)(n-j+1)} $.
This comes from my answer here: Cauchy product of $\sum\limits_n^{\infty}\frac{1}{n}$ with itself
Numerical experimentation suggests that $g(n)$ is decreasing and is about $\dfrac1{\sqrt{n}-1}$, so it seems that $\lim_{n \to \infty} \sqrt{n}g(n) $ exists and might be $1$.
(I wouldn't be surprised if this has already been answered.)
Note that
$$\begin{align} \sum_{j=0}^n\left(\frac{1}{(j+1)(n-j+1)}\right)&=\frac{1}{n+2}\sum_{j=0}^n \left(\frac{1}{j+1}+\frac{1}{n-j+1}\right)\\\\ &=\frac{2}{n+2}\sum_{j=0}^n\frac{1}{j+1}\\\\ &=\frac{2}{n+2}H_{n+1} \end{align}$$
Hence, as $n\to \infty$, we see that
$$\begin{align} \sum_{j=0}^n\left(\frac{1}{(j+1)(n-j+1)}\right)&=\frac{2\log(n+1)}{n+2}+\frac{2\gamma}{n+2}+O\left(\frac{1}{n^2}\right)\\\\ &=\frac{2\log(n)}{n}+\frac{2\gamma}{n}+O\left(\frac1{n^2}\right) \end{align}$$