Problem w/ limit which I do not understand: lim[x->+inf] (1-1/2)*(1-1/3)*...*(1-1/n)

[wolly]

Problem w/ limit which I do not understand: lim[x->+inf] (1-1/2)*(1-1/3)*...*(1-1/n)

lim (1-1/2)*(1-1/3)*.....*(1-1/n) and the answer would be 0.I searched my textbook and it showed me this 1/n.How did it get there?I searched google for answers and I got (n-2)*(n-1)/(n-1)/n which is x->+inf n-2/n but no proof at all and after that they get 2/n.Could someone help me?

 

[wolly]

Limit problem

How do I solve this?
lim (1-1/2)*(1-1/3)*...*(1-1/x)=0 Also in my textbook I got the
x->+inf answer 1/x.
Do I solve it like thisn-1)!/n!?

 

[Dr.Peterson]

How do I solve this?
lim (1-1/2)*(1-1/3)*...*(1-1/x)=0 Also in my textbook I got the answer 1/x.
x->+inf
Do I solve it like this: (n-1)!/n!?

That would do it.

Do you see why? What is it that you need help with?

 

[mmm4444bot]

… a limit which I do not understand …

It's hard to know where to help because you haven't shown what you tried yourself. Did you write out some factors, to see what's happening?

The factors are 1 - 1/n, for Natural numbers n. I'd start by doing that subtraction (i.e., get a single ratio).

1 - 1/n = (n - 1)/n

This shows each factor is a ratio and its numerator is always one less than its denominator. As an example, let's say n=5. The nth ratio will have 5 as its denominator.

\(\displaystyle \displaystyle \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5}\)

If you were to multiply these ratios by hand, you would begin by cancelling the first three denominators with the corresponding numerators, yes? Write it out on paper, and do it.

Pick two other values of n, and write out the ratios and determine each product. Hopefully, writing out examples will help you see why the product is always the same expression (in terms of n). :cool: