Infinite Sum

[Guest]

Calculate the sum:

[tex](1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}).......(1-\frac{1}{n})[/tex]

[Guest]

By evaluating the brackets
[tex]\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdots \left(1-\frac{1}{n-1}\right)\left(1-\frac{1}{n}\right)[/tex]
becomes
[tex]\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdots\frac{n-2}{n-1}\cdot\frac{n-1}{n}[/tex]
The numerators of the fractions cancel with the denominators of the preceding fraction, leaving the answer
[tex]\frac{1}{n}[/tex]

Hope this helped,

R. Baber.

[Guest]

Thank you Mr. Baber.

[Guest]

It is a finite product not an ifinite sum but the method of evaluating it is correct.

[Guest]

I meant finite product not an infinite sum.

[Guest]

Yeah that is right. It is an infinite product.

[leesajohnson]

Infinite Sum ?