# Does #a_n=(8000n)/(.0001n^2) #converge? If so what is the limit?

##### 1 Answer

The sequence converges to 0.

#### Explanation:

We can evaluate the limit of this series by looking at dominant terms. The dominant term in the numerator is

#lim_(n->oo)(8000n)/(0.0001n^2)#

#=>lim_(n->oo)n/n^2#

#=>lim_(n->oo)1/n#

As

#=>lim_(n->oo)1/n=0#

Alternatively, you can use L'Hospital's rule to simplify, though this takes a bit more work. It will yield the same result. By this rule, if:

#lim(f(x))/(g(x))=0/0# or#(+-oo)/(+-oo)# , then#lim(f(x))/(g(x))=lim(f'(x))/(g'(x))#

#lim_(n->oo)(8000n)/(0.0001n^2)=>(oo)/(oo)#

Take the derivative of the numerator and denominator:

#lim_(n->oo)=(8000)/(0.0002n)#

As