Posted by **Hannah** on Tuesday, July 7, 2009 at 8:15am.

a sequence {an} is given by a1 = (3)^1/2 , an+1 = (3+an)^1/2. Show that {an} is monotonic and bounded, and calculate the limit if it exists

## Answer This Question

## Related Questions

- Math - A sequence {an} is given by a1=root(3), a n+1 = root(3+an). show that {an...
- calculus - A sequence{an} is given by a1=sqrt(2), an+1=sqrt(2*an). a) by ...
- Limit of Sequence - I need help finding the limit of the following sequence or ...
- Calculus - Determine whether the sequence with the given nth term is monotonic. ...
- Math - (a) Find the number c such that the limit below exists. Limit as x goes ...
- Math - limit of (x,y)--->(1,0) of ln(1+y^2/x^2+xy)) Find the limit, if it ...
- Calc 2 - I need help finding the limit of the following sequence or determining ...
- Calculus - limit of (x*(y-1)^2*cosx)/(x^2+2(y-1)^2) as (x,y)->(0,1). By ...
- calculus - (a) Find the number c such that the limit below exists. Limit as x ...
- trig - find the limit of the sequence if it exists tn= 6/n

More Related Questions