Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$ I am trying to show that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000}
=\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}.
$$
It seems there is some sort of generalizable pattern here, so I will verify it for smaller numbers:
$$
\begin{align*}
\text{Say, }n=4 \hspace{35pt} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}&=\frac{1}{3}+\frac{1}{4}\\
\frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}&=\frac{4}{12}+\frac{3}{12}\\
\frac{7}{12}&=\frac{7}{12}
\end{align*}
$$
So, my guess on the general formula is
$$
 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n/2+1}+\ldots+\frac{1}{n-1}+\frac{1}{n}.
$$
This really seems like I am getting somewhere, but how can I finish off the proof? Is induction viable?

Let $$S(n) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n}.$$
Then notice if we add $$T(n) = 2 \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n}\right)$$ to $S(n)$, all the negative terms become positive and we get a nice sum:
$$S(n) + T(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n} = H(2n),$$ where $H(n) = 1 + 1/2 + \cdots + 1/n$ is the $n^{\rm th}$ harmonic number.
But notice that $T(n)$ is itself a harmonic number:  just distribute the $2$:  $$T(n) = \frac{2}{2} + \frac{2}{4} + \cdots + \frac{2}{2n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} = H(n),$$ so they are actually the same.  Therefore,
$$S(n) = H(2n) - H(n).$$  And from here it is easy to see that
$$S(n) = \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n}\right) - \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) \\
= \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n},$$
which proves the claim.
