Help with this Telescoping Series

[Je m'appelle]

Homework Statement

Show that

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} = 1$$

Homework Equations

The Attempt at a Solution

So as this is a telescoping series, I rewrote the general formula through partial fractions as

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)} - \frac{1}{(n+2)} = 1$$

The first few terms will be

$$(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{7}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+2)})$$

It can be seen then, that the first term $$\frac{1}{2}$$ and the last term $$\frac{-1}{(n+2)}$$ do not cancel, therefore, it turns out to be basically

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)} - \frac{1}{(n+2)} = \lim_{n \rightarrow \infty} (\frac{1}{2} - \frac{1}{(n+2)}) = \frac{1}{2}$$

What am I doing wrong here? Is this telescoping series equal to 1 or 1/2? My textbook says 1 as I pointed out in the beginning of this thread, but I came up with 1/2, so I'd like to know if I'm doing something wrong or if I'm correct and the textbook has a typo.

Thanks in advance.

 

[statdad]

You haven't done anything wrong - the sum doesn't equal 1.

 

[tiny-tim]

Hi Je m'appelle!

Your result looks correct to me.

The given answer would be for a ∑ starting at n = 0, not n = 1.

 

[Je m'appelle]

You haven't done anything wrong - the sum doesn't equal 1.

Hi Je m'appelle!

Your result looks correct to me.

The given answer would be for a ∑ starting at n = 0, not n = 1.

Yes, it seems then that it is a typo of either the sum being equal to 1/2 or n being equal to 0 heh instead of 1 and 1 respectively.

Thank you for your help statdad and tiny-tim