Find an expression, in terms of n, for $\sum_{r=2}^n {1\over r-1}-{1\over r+1} $ Working:
$$\sum_{r=2}^n {1\over r-1}-\sum_{r=2}^n {1\over r+1} $$
$$={1\over1} + {1\over2} + {1\over3}+\cdots+{1\over n-1}$$
$$-\left({1\over3}+\cdots+{1\over n-1}+{1\over n+1}\right)$$
This should then make:
$$1+{1\over2}-{1\over n+1}$$
But it is:
$$1+{1\over2}-{1\over n}-{1\over n+1}$$
Why is there the $-{1\over n}$ term?
Also how do you solve $\sum_{r=1000}^\infty {1\over r-1}-{1\over r+1}  $  I haven't the faintest.
Thanks

We want to find a simple expression for
$$\sum_{r=2}^n \left(\frac{1}{r-1}-\frac{1}{r+1}\right).$$
To get an idea about your lost $-\dfrac{1}{n}$, let us add a few terms together, maybe up to $r=7$, to see what's going on. We get
$$\left(1-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right) +\left(\frac{1}{5}-\frac{1}{7}\right)+ \left(\frac{1}{6}-\frac{1}{8}\right) .$$
There is a lot of cancellation. Everything disappears except for the $1$, the $\dfrac{1}{2}$, the $-\dfrac{1}{7}$, and the $-\dfrac{1}{8}$.
The $-\dfrac{1}{7}$ is your missing $-\dfrac{1}{n}$ term. 
Added: The OP has been expanded to include the reasoning. Somewhat altered, it says that the sum is equal to
$$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-2}+\frac{1}{n-1}     \right)-\left(\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}     \right).$$
Everything cancels except the first two terms and the last two.
