How find this sum $\sum_{k=0}^{\left[\frac{n}{2}\right]}\frac{(-1)^k\binom{n-k}{k}}{n-k}$ How  find this sum 
$$\sum_{k=0}^{\left[\dfrac{n}{2}\right]}\dfrac{(-1)^k\binom{n-k}{k}}{n-k}$$
My try:since 
$$\dfrac{(-1)^k}{n-k}=\int_{-1}^{0}x^{n-k-1}dx$$
then I can't 
Thank you very much!

Following the well-known  technique by Wilf we  introduce the ordinary
generating function $f(z) = \sum_{n\ge 1} f_n z^n$ where $f_n$ is our 
sum so that
$$ f(z) = \sum_{n\ge 1} z^n 
\sum_{k\ge 0} (-1)^k {n-k \choose k} \frac{1}{n-k}$$
and re-arrange terms to get
$$ f(z) = \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k 
\sum_{n\ge 2k} {n-k\choose k} \frac{z^n}{n-k}$$
which is
$$ f(z) = \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k 
\sum_{m\ge 0} {m+k\choose k} \frac{z^{2k+m}}{m+k}
\\=   \log\frac{1}{1-z} + \sum_{k\ge 1} (-1)^k z^{2k}
\sum_{m\ge 0} \frac{m+k}{k} {m+k-1\choose k-1} \frac{z^m}{m+k}
\\=  \log\frac{1}{1-z} + \sum_{k\ge 1} (-1)^k \frac{z^{2k}}{k}
\sum_{m\ge 0} {m+k-1\choose k-1} z^m
\\ =  \log\frac{1}{1-z} +
\sum_{k\ge 1} (-1)^k \frac{z^{2k}}{k} \frac{1}{(1-z)^k}$$
which finally becomes
$$  \log\frac{1}{1-z} + \log \frac{1}{1+z^2/(1-z)}
=  \log\frac{1}{1-z} + \log \frac{1-z}{1-z+z^2}
= \log \frac{1}{1-z+z^2}.$$
Now the roots of the denominator of the fractional term are
$$\rho_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}i}{2}$$
and in particular we observe that $$1-z+z^2 = (1-z/\rho_1)(1-z/\rho_2)$$
(which is not true for the general quadratic), so we have
$$f(z)
= 
- \log(1-z/\rho_1) 
- \log(1-z/\rho_2)
= \log\frac{1}{1-z/\rho_1} + \log\frac{1}{1-z/\rho_2}.$$
Extracting  coefficients from  this  formula we  obtain the  following
exact expression for $f_n:$
$$f_n = 
\frac{\rho_1^{-n}}{n} + \frac{\rho_2^{-n}}{n}
= \frac{\rho_1^n}{n} + \frac{\rho_2^n}{n}$$
because $\rho_1\rho_2 = 1.$
Here we have made repeated use of the following expansion which is popular in combinatorics because it is the generating function of the species of cycles:
$$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}.$$ 
Two similar calculations can be found at this MSE link.
