A little problem with a binomial identity I have to compute the quantity $\sum\limits_{k=0}^{n-1} \binom{n}{k} \frac{k}{n-k}$.
Using the identity $k\binom{n}{k}=n\binom{n-1}{k-1}$ and reindexing the sum, it's easy to see the previous sum becomes
$$
\sum\limits_{j=0}^{n-2} \binom{n-1}{j} \frac{1}{(n-j-1)^2}
$$
But now I don't know how to continue. Can you help me?

$$\begin{eqnarray*}\sum_{k=0}^{n-1}\binom{n}{k}\frac{k}{n-k}&=&\sum_{k=1}^{n}\binom{n}{k}\frac{n-k}{k}=(2^n-1)+n\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}\\&=&(2^n-1)+n\int_{0}^{1}\frac{(1+x)^n-1}{x}\,dx\\&=&(2^n-1)+n\sum_{k=1}^{n}\frac{2^{k}-1}{k}\end{eqnarray*}$$
is easier to bound but cannot be simplified much further.
