How find this sum $ \frac{1}{1999}\binom{1999}{0}-\frac{1}{1998}\binom{1998}{1}+\cdots-\frac{1}{1000}\binom{1000}{999}$ 
prove or disprove :
$$S=\dfrac{1}{1999}\binom{1999}{0}-\dfrac{1}{1998}\binom{1998}{1}+\dfrac{1}{1997}\binom{1997}{2}-\dfrac{1}{1996}\binom{1996}{3}+\cdots-\dfrac{1}{1000}\binom{1000}{999}=\dfrac{1}{1999}\left(w^{1999}_{1}+w^{1999}_{2}\right)$$
where
$$w_{1}=\dfrac{1+\sqrt{3}i}{2},w_{2}=\dfrac{1-\sqrt{3}i}{2}$$

My idea: since
$$\dfrac{1}{1999-k}\binom{1999-k}{k}=\dfrac{1}{1999-k}\dfrac{(1999-k)!}{k!(1999-2k)!}=\dfrac{(1998-k)!}{k!(1999-2k)!}$$
Then I can't,maybe can use integral deal it
Thank you
 A: The Chebyshev polynomials of the first kind are given by
\begin{align}
T_{n}(x) &= \frac{n}{2} \sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} (2x)^{n-2k} \\
&= \frac{1}{2} \left[ (x - \sqrt{x^{2}-1})^{n} + (x - \sqrt{x^{2}-1})^{n} \right].
\end{align}
When $x=1/2$ it is seen that
\begin{align}
\sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} = \frac{1}{n} \left( a^{n} + b^{n} \right)
\end{align}
where $2a = 1+\sqrt{3} i$ and $2b=1-\sqrt{3} i$. When $n=1999$ this becomes
\begin{align}
\sum_{k=0}^{999} \frac{(-1)^{k}}{1999-k} \binom{1999-k}{k} = \frac{1}{1999} \left( a^{n} + b^{n} \right).
\end{align}
Thus the relationship is shown to be true.
If the sum is changed to all positive terms it can easily be seen as a Lucas number, namely,
\begin{align}
\sum_{k=0}^{999} \frac{1}{1999-k} \binom{1999-k}{k} = \frac{L_{1999}}{1999}.
\end{align}
A: This sum can also be done using Wilf / generatingfunctionology. Let
$$a_n = \sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k}$$
and introduce the generating function
$$f(z) = \sum_{n\ge 1} a_n z^n
= \sum_{n\ge 1}
\sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n
\\ =
\sum_{n\ge 1} \frac{z^n}{n} +
\sum_{n\ge 1}
\sum_{k=1}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n.$$
Switch the order of summation in the inner sum to obtain
$$\sum_{k\ge 1} \sum_{n\ge 2k} 
\frac{(-1)^k}{n-k} {n-k \choose k} z^n
= \sum_{k\ge 1} \sum_{n\ge 0} 
\frac{(-1)^k}{n+k} {n+k \choose k} z^{n+2k}
\\= \sum_{k\ge 1} z^{2k} \sum_{n\ge 0} 
\frac{(-1)^k}{n+k} {n+k \choose k} z^n
= \sum_{k\ge 1} z^{2k} \sum_{n\ge 0} 
\frac{(-1)^k}{k} {n+k-1 \choose k-1} z^n
\\ = \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \sum_{n\ge 0} 
{n+k-1 \choose k-1} z^n
= \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \frac{1}{(1-z)^k}
\\ = \log\frac{1}{1 + \frac{z^2}{1-z}}.$$
This gives the following closed form for $f(z):$
$$f(z) = \log\frac{1}{1-z} +
\log\frac{1}{1 + \frac{z^2}{1-z}}
= \log\frac{1}{1-z+z^2}.$$
Finally put
$$w_{1,2} = \frac{1\pm i\sqrt{3}}{2}$$
so that $$1-z+z^2
= (1-w_1 z)(1-w_2 z).$$
This yields
$$f(z) = \log\frac{1}{1-z+z^2} =
\log\frac{1}{1-w_1 z} + \log\frac{1}{1-w_2 z}$$
so that
$$[z^n] f(z) = \frac{w_1^n}{n} + \frac{w_2^n}{n},$$
as obtained by @r9m in the comment.
Remark. Here we have made repeated use of the identity
$$\log\frac{1}{1-z} = \sum_{n\ge 1} \frac{z^n}{n}.$$
