Approximation of sum $\sum_{j=1}^{n-1}\frac{2j}{(n-j+1)(j+n)}$ I'm trying to solve this sum or at least give an approximation.
$$\sum_{j=1}^{n-1}\frac{2j}{(n-j+1)(j+n)}$$
This was my attempt:
Assuming that in my sum $\sum f(j)$, $f(j)$ is a monotonically increasing function, then I can approximate the sum by the following rule:
$$\int_{m-1}^{n} f(x) dx \leq \sum_{j=m}^{n} f(j) \leq \int_{m}^{n+1} f(x)dx$$ 
So I need to solve this integral:
$$\sum_{j=1}^{n-1}\frac{2j}{(n-j+1)(j+n)} \leq \int_{1}^{n} \frac{2j}{(n-j+1)(j+n)}dj = \frac{2(\ln n+n \ln (\frac{n+1}{2}))}{2n+1}$$
Then since,
$$\frac{2(\ln n+n\ln(\frac{n+1}{2}))}{2n+1} = \frac{2 \ln n}{2n + 1} +  \frac{2 n \ln (\frac{n+1}{2})}{2n + 1}$$
The upper bound for the both terms is
$$O\left(\frac{\ln n}{n}\right) + O(\ln n) = O(\ln n)$$
Then if my calculations are correct I should can prove by induction that, the following inequality holds, for some constant $c$.
$$\sum_{j=1}^{n-1}\frac{2j}{(n-j+1)(j+n)} \leq c\ln(n)$$
But I don't know how to prove it.
 A: Let $$s_n=\sum_{j=1}^{n-1}\frac{2j}{(n-j+1)(j+n)}$$ Now, define
$$a_n=\sum_{j=1}^{n-1}\frac{1}{n-j+1}$$
and $$b_n=\sum_{j=1}^{n-1}\frac{1}{j+n}$$
Then $$\begin{split}
a_n-b_n &= \sum_{j=1}^{n-1}\bigg(\frac{1}{n-j+1}-\frac{1}{j+n}\bigg)\\
&= \sum_{j=1}^{n-1}\frac{2j-1}{(n-j+1)(j+n)}\\
&= s_n-\sum_{j=1}^{n-1}\frac{1}{(n-j+1)(j+n)}
\end{split}
$$
Also
$$\begin{split}
a_n+b_n &= \sum_{j=1}^{n-1}\bigg(\frac{1}{n-j+1}+\frac{1}{j+n}\bigg)\\
&= \sum_{j=1}^{n-1}\frac{2n+1}{(n-j+1)(j+n)}\\
\end{split}
$$
Therefore
$$a_n-b_n=s_n-\frac{a_n+b_n}{2n+1}$$
In other words $$s_n=\frac{2n+2}{2n+1}a_n-\frac{2n}{2n+1}b_n$$
Now we need to estimate $a_n$ and $b_n$.
Changing $j$ into $n-j+1$ in the definition of $a_n$ yields
$$a_n=\sum_{j=2}^{n}\frac1j$$
So $a_n \approx \ln n$ (see harmonic series).
For $b_n$, note that it's a Riemann sum:
$$b_n=\frac 1 n \sum_{j=1}^{n-1}\frac{1}{\frac j n +1}$$
therefore $$b_n\approx \int_0^1\frac {dx}{1+x}=\ln 2$$
We conclude that $$s_n\approx \ln n$$
A: Hint 
$\dfrac{2j-1}{(n-j+1)(j+n)} = \dfrac{1}{n+j}-\dfrac{1}{n-j+1}$ and $\dfrac{(2n+1)}{(n-j+1)(n+j)} = \dfrac{1}{n+j} +\dfrac{1}{n-j+1}$ 
So you have : $\sum \frac{1}{n+j} - \sum\frac{1}{n-j+1} + \frac{1}{(2n+1)}\left( \sum \frac{1}{n+j} + \sum \frac{1}{n-j+1}\right)$~$\log(2n) - 2\log(n) + \frac{1}{2n+1}\log(2n)$
A: With CAS help:
$$\color{red}{\sum _{j=1}^{n-1} \frac{2 j}{(n-j+1) (j+n)}}=\\\sum _{j=1}^{n-1} -\frac{2 (1+n)}{(-1+j-n) (1+2 n)}+\sum _{j=1}^{n-1} -\frac{2 n}{(j+n) (1+2 n)}=\\\frac{(-2-2
   n) (1-\gamma )}{1+2 n}-\frac{(-2-2 n) \psi (1+n)}{1+2 n}-\frac{2 (n \psi (2 n)-n \psi (1+n))}{1+2 n}=\color{red}{\\-\frac{2}{1+2 n}-\frac{2 n}{1+2 n}-\frac{2 \gamma
    n}{1+2 n}+\frac{2 H_n}{1+2 n}+\frac{4 n H_n}{1+2 n}-\frac{2 n \psi (2 n)}{1+2 n}}$$
where:$H_n$ is Harmonic number, $\psi$ is digamma function and $\gamma$ is Euler gamma constant.
Mathematica code:
 Sum[(2*j)/((n - j + 1)*(j + n)), {j, 1, n - 1}] == 
 -(2/(1 + 2 n)) - (2 n)/(1 + 2 n) - (2 EulerGamma n)/(1 + 2 n) + (
 2 HarmonicNumber[n])/(1 + 2 n) + (4 n HarmonicNumber[n])/(
 1 + 2 n) - (2 n PolyGamma[2 n])/(1 + 2 n)

