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?

above, both are excellent answers. This is another way to look at it.
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$
After cancelling out positive matching terms on both sides and moving negative matching terms to RHS, we get:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{999}-\frac{1}{1000} =\frac{1}{501}+\frac{1}{502}+\ldots+\frac{1}{999}+\frac{1}{1000}$
Continuing the descent, we will end up with:
$1-\frac{1}{2}+\frac{1}{3}=\frac{1}{2}+\frac{1}{3}$
which is true.
