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?
 A: 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.
A: Proceed with induction. We seek to show, in essence, for all $n$ sufficiently large,
$$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$
We have a base case. So let's try to show that - assuming the above holds for $n$ - that it holds for $n+1$ too. We have
$$\begin{align*}
\sum_{k=1}^{2(n+1)} \frac{(-1)^{k-1}}{k}
&= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2} + \sum_{k=1}^{2n} \frac{(-1)^k}{k} \\
&= \frac{1}{2n+1} + \frac{-1}{2n+2} + \sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} \\
&= \frac{1}{2n+1} + \frac{-1}{2n+2} + \sum_{k=n+1}^{2n} \frac{1}{k} \\
\sum_{k=(n+1)+1}^{2(n+1)} \frac{1}{k}
&= \sum_{k=n+2}^{2n+2} \frac{1}{k} \\
&= \frac{1}{2n+2} + \frac{1}{2n+1}+ \sum_{k=n+2}^{2n} \frac{1}{k} \\
&= \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1} + \sum_{k=n+1}^{2n} \frac{1}{k} \\
\end{align*}$$
It is trivial to show that
$$\frac{1}{2n+1} + \frac{-1}{2n+2} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$
so the induction follows and the original equation holds for $n \ge 4$. In particular, $n=1000$ gives you your desired result.
A: 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.
