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 $0<c<1$? Can we calculate $S(n,c)$ exactly? What about asymptotics? Can we find upper/lower bounds?
 A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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When $c = 1$, $\color{#0000ff}{\large{\rm S}\pars{n,1}} = \braces{\pars{n - 1}\bracks{\pars{n - 1} + 1}/2}/n = \color{#0000ff}{\large\pars{n - 1}/2}$.
Let's consider the case $c \not= 1$:
\begin{align}
{\rm S}\pars{n, c} &= \sum_{i = 1}^{n - 1}{i \over ci + \pars{n-i}}
= 
\sum_{i = 1}^{n - 1}{i \over \pars{c - 1}i  + n}
=
{1 \over c- 1}\sum_{i = 1}^{n - 1}{i \over i  + n/\pars{c - 1}}
\\[3mm]&=
{1 \over c- 1}\sum_{i = 1}^{n - 1}
\bracks{1 - {n/\pars{c - 1} \over i  + n/\pars{c - 1}}}
={n - 1 \over c - 1} - {n \over \pars{c -1}^{2}}\sum_{i = 0}^{n - 2}
{1 \over i + n/\pars{c - 1} + 1}
\\[3mm]&=
{n - 1 \over c - 1} - {n \over \pars{c - 1}^{2}}
\braces{\Psi\pars{\bracks{{n \over c - 1} + 1} + n - 1} - \Psi\pars{{n \over c - 1} + 1}}
\end{align}
\begin{align}
\color{#0000ff}{\large{\rm S}\pars{n, c \not= 1}}
&= \sum_{i = 1}^{n - 1}{i \over ci + \pars{n-i}}
\\[3mm]&=\color{#0000ff}{\large{n \over c - 1} + {n \over \pars{c - 1}^{2}}
\bracks{\Psi\pars{{n \over c - 1}} - \Psi\pars{n\,{c \over c- 1}}}}
\end{align}
$\Psi\pars{z}$ is the $\it digamma$ function and we used the identities:
$$
\Psi\pars{x + m} = \Psi\pars{x} + \sum_{k = 0}^{m - 1}{1 \over x + k}\,,\qquad
\Psi\pars{1 + z} = \Psi\pars{z} + {1 \over z}
$$

When $0 < c < 1$, the digamma functions arguments go to $-\infty$. It's convenient to use the $\it\mbox{digamma reflexion formula}$:
$$
\Psi\pars{z} = \Psi\pars{1 - z} - \pi\cot\pars{\pi z}
$$
$$\left\lbrace%
\begin{array}{rcl}
\Psi\pars{n \over c - 1}
& = &
\Psi\pars{1 + {n \over 1 - c}} + \pi\cot\pars{\pi n \over 1 - c}
=
\Psi\pars{n \over 1 - c} + {1 - c \over n} + \pi\cot\pars{\pi n \over 1 - c}
\\[1mm]
\Psi\pars{nc \over c - 1}
& = &
\Psi\pars{1 + {nc \over 1 - c}} + \pi\cot\pars{\pi nc \over 1 - c}
=
\Psi\pars{nc \over 1 - c} + {1 - c \over nc} + \pi\cot\pars{\pi nc \over 1 - c}
\end{array}\right.
$$
and
\begin{align}
{\rm S}\pars{n,c}
&=
-\,{1 \over c\pars{1 - c}}
\\[3mm]&+ {n \over \pars{c - 1}^{2}}\bracks{%
\Psi\pars{n \over 1 -c} - \Psi\pars{nc \over 1 - c}
+ \pi\cot\pars{\pi n \over 1 -c} - \pi\cot\pars{\pi nc \over 1 -c}}
\\[3mm]
\left.\vphantom{\LARGE A}{\rm S}\pars{n,c}\right\vert_{0\ <\ c\ < 1 \atop n\ \gg\ 1}
&\sim
-\,{1 \over c\pars{1 - c}} + {n \over \pars{c - 1}^{2}}\bracks{%
\ln\pars{1 \over c} + \pi\cot\pars{\pi n \over 1 -c}
-
\pi\cot\pars{\pi nc \over 1 -c}}
\end{align}
where we used the digamma function $\it\mbox{asymptotic behavior}$:
$\Psi\pars{z} \sim \ln\pars{z}$ when $\verts{z} \gg 1$. Notice that the $\cot$'s terms are "oscillating ones".

Notice that
$$
\pi\cot\pars{\pi n \over 1 -c} - \pi\cot\pars{\pi nc \over 1 -c}
=
-\,{\pi\sin\pars{n\pi}
    \over \sin\pars{\pi n/\bracks{1 -c}}\sin\pars{\pi nc/\bracks{1 -c}}}
$$
and it vanishes out whenever
$n/\pars{1 - c}\ \mbox{and}\ nc/\pars{1 - c}\ \not\in {\mathbb N}$.
A: *

*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.
