Calculate the sum

[Math Tutor]

Calculate the sum:
[tex]\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{3\cdot2}+...+\frac{1}{n(n+1)}[/tex]

[Guest]

Are you sure you typed the question correctly? I don't see the pattern of your terms.

The start of your sum are the first three values of the sequence [tex]\frac{1}{n\cdot 2}[/tex] which doesn't match the last term [tex]\frac{1}{n(n+1)}[/tex].

If you meant [tex]\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{n(n+1)}[/tex]
the answer is given below

Spoiler: show

[tex]\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}[/tex]
so
[tex]\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{n(n+1)} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \ldots + (\frac{1}{n} - \frac{1}{n+1}) = 1 - \frac{1}{n+1}[/tex]

R. Baber

[Math Tutor]

Perfect!