Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.