I wish to find a closed form, or a good upper bound for $\sum_{i=1}^{i=n-1} \frac{1}{i \times (n-i)}$.
I can specify a lower bound of $(n-1)/n^2$, which looks like $1/n$ because we have $n-1$ terms are greater than $1/n^2$, and an upper bound in the neighborhood $log(n)$ due to individual terms being less than $1/i$, and our knowledge that sums of $1/i$ approximate the log function but that's about it.
Note that this is also the convolution of $1/n$ with itself, if this helps.
Requested in comments:
User:Ramanujan suggested dividing up the summand into partial fractions.
You found $\frac{1}{i (n-i)} = \frac{1}{ni} + \frac{1}{n (n-i)}$ (reversing two signs since $n>i$) and so related to the sum of two harmonic series, and suggested the reasonable approximation $2\log(n)/n.$
Since the sum is exactly $\frac2nH_{n-1}$, you could find tighter bounds and say it is between $\dfrac{2}n(\log (n-1)+\gamma)+\dfrac1{n(n-1)}-\dfrac1{4n(n-1)^2}$ and $\dfrac{2}n(\log (n-1)+\gamma)+\dfrac1{n(n-1)},$ where $\gamma\approx 0.5772156649$ is the Euler–Mascheroni constant.