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

Question

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

Scarlett Allison · Accepted Answer

The sequence converges to 0.

We can evaluate the limit of this series by looking at dominant terms. The dominant term in the numerator is #n#, while the dominant term in the denominator is #n^2#. #lim_(n->oo)(8000n)/(0.0001n^2)# #=>lim_(n->oo)n/n^2# #=>lim_(n->oo)1/n# As #n# approaches infinity, #1/n# approaches #0#. #=>lim_(n->oo)1/n=0# #:.# The sequence converges to 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 #n# approaches infinity, #8000/(0.0002n)# approaches #0#. Essentially, you are dividing #8000# by bigger and bigger numbers until the number in the denominator is so large that the quotient is #~~0#.

Adam Adkins · Answer

undefined To determine if the sequence $ a_n = \frac{8000n}{0.0001n^2} $ converges, we can analyze its behavior as $ n $ approaches infinity.

First, simplify the expression:

$ a_n = \frac{8000n}{0.0001n^2} = \frac{8000}{0.0001n} $

As $ n $ approaches infinity, the denominator $ 0.0001n $ becomes infinitely large, and the sequence tends towards zero.

Therefore, the sequence converges, and its limit is $ 0 $.