# Explaining the pattern

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} =\frac12 + \frac16 = \frac36+\frac16 = \frac46 = \boxed{\tfrac23}$$

$$\left(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}\right) +\frac{1}{3\cdot 4} =\frac23 +\frac{1}{3\cdot 4} = \frac{2\cdot4}{3\cdot 4} + \frac{1}{3\cdot 4} =\frac{9}{12} = \boxed{\dfrac34}$$

Every time I add a fraction whose numerator is 1, and whose denominator is the product of the first term of the denominator of the previous fraction + 1 x the second term of the denominator of the previous fraction If I keep going I'll have $$\frac{4}{5}$$, then $$\frac{4}{5}$$ and so on. Why is there this pattern?

Because $$\frac{1}{n(n+1)} = \frac1n - \frac{1}{n+1}$$.

$$\frac{1}{n\cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots + \frac{1}{n\cdot (n+1)}$$

$$= \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots +\frac{1}{n}-\frac{1}{n+1}$$

Terms in the middle cancels out

$$=1-\frac{1}{n+1}=\frac{n}{n+1}$$

Which is what you observed.

• I still don't get it, as far as the first equation you've written is concerned, why is it true?
– Pier
May 22, 2020 at 13:06
• @Pier $\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}$ can be proven by the reveresed direction. To "intuitively see how to find the RHS", look up "partial fraction decomposition" May 22, 2020 at 13:09