Sum $S(n,c) = \sum_{i=1}^{n-1}\dfrac{i}{ci+(n-i)}$ Consider the sum $$S(n,c) = \sum_{i=1}^{n-1}\dfrac{i}{ci+(n-i)}$$ where $0\le c\le 1$. 

When $c=0$, $S(n,c)$ grows asymptotically as $n\log n$.

When $c=1$, $S(n,c)$ grows asymptotically as $n$.

What about when $0&lt;c&lt;1$? Can we calculate $S(n,c)$ exactly? What about asymptotics? Can we find upper/lower bounds?
We have
$$S(n;c) = \sum_{k=1}^{n-1} \dfrac{k}{ck+(n-k)} = \sum_{k=1}^{n-1} \dfrac{k/n}{1+(c-1)k/n}$$
Hence,
$$\dfrac{S(n;c)}n = \sum_{k=1}^{n-1} \dfrac{k/n}{1+(c-1)k/n} \dfrac1n \sim \int_0^1 \dfrac{xdx}{1+(c-1)x} = \dfrac{(c-1)-\log(c)}{(c-1)^2} \text{ for }c \in (0,1)$$
Hence,
$$S(n;c) \sim \dfrac{(c-1)-\log(c)}{(c-1)^2} n \text{ for }c \in (0,1)$$
Better approximations can be obtained using Euler Maclaurin formula.
There is a closed form, thanks to mathematica (using digammas):


$\frac{c n+n \psi ^{(0)}\left(\frac{c}{c-1}+\frac{n}{c-1}-\frac{1}{c-1}\right)-n
   \psi ^{(0)}\left(\frac{n
   c}{c-1}+\frac{c}{c-1}-\frac{1}{c-1}\right)-n}{(c-1)^2}$


There are, therefore, asymptotics (also thanks to mathematica), on which MathJax seems to choke, here is a link to the screen grab:
https://www.evernote.com/shard/s24/sh/e1d1abb1-b8a4-4f4f-bad8-7e476f73f017/7700044225f8b6d0d008346a44a2fc5d


These can be simplified, if one knew the sign of $c,$ but I leave this to you.