How do you do the limit comparison test for this problem #sqrt ( (n+1)/ (n^2+2))# as n goes to infinity?

Question

Paisley Johnson · Accepted Answer

Diverges when compared to #b_n=1/sqrtn#

We need to come up with a new sequence #b_n# to compare to #a_n=sqrt((n+1)/(n^2+2))#. Furthermore, we need to be able to easily determine the divergence of #sum_(n=1)^oob_n#. Generally, we'll use the #p-#series test.

So, let's create #b_n# by ignoring the constant values of #1,2# in the original series.

#b_n=sqrt(n/n^2)=sqrt(1/n)=1/sqrtn=1/n^(1/2)#

Now, we know #sum_(n=1)^oo1/n^(1/2)# diverges by the #p-#series test, #p=1/2<1#.

The Limit Comparison tells us if we know the convergence or divergence of #a_n# or #b_n# and

#c=lim_(n->oo)a_n/b_n>0ne+-oo#, then both series either converge or diverge.

Knowing #a_n=sqrt((n+1)/(n^2+2)), b_n=1/sqrtn, sum_(n=1)^oob_n# diverges,

#c=lim_(n->oo)sqrt((n+1)/(n^2+2))/(1/sqrtn)=lim_(n->oo)sqrt((n(n+1))/(n^2+2))=lim_(n->oo)sqrt((n^2+n)/(n^2+2))=1>0ne+-oo#

Then, both series diverge.

Christopher Agresta · Answer

The series diverges.

See work below:

The idea of the limit comparison test is that you essentially compare your unknown function to a function whose convergence you know (through another method: typically p-test). Here's how you do this: Let's say the series you want to analyze is #a_n#. We pick a similar series of known convergence (call that #b_n#), and do the following: #lim_(n->oo) a_n/b_n# #a_n >= 0, b_n > 0# for all #n#. If this limit #color(red)(= 0)#, then #color(red)("both series converge")# If this limit #color(green)(= L)# (any arbitrary nonzero value), then #color(green)(a_n " does what " b_n " does.")# If this limit #color(blue)("goes to infinity")#, then #color(blue)("both series diverge")#. So now, we figure out what series it would be ideal to compare this to. I'm going to chose #sqrt(n/n^2)# which, with some simplification, turns into #1/sqrt(n)#. By the p-test, we know that this series diverges. Bearing this in mind, let's take the limit: #=> lim_(n->oo) sqrt ( (n+1)/ (n^2+2))/(1/sqrt(n))# #=> lim_(n->oo) sqrt((n^2 + n)/(n^2+2))# Now, we just evaluate this limit using the same steps we learned in Calc 1. We just divide every term by the highest power: #=> lim_(n->oo) sqrt((n^2/n^2 + n/n^2)/(n^2/n^2+2/n^2))# #=> lim_(n->oo) sqrt((1 + 1/n)/(1+2/n^2))# ..and now take the limit as #n->oo# of each term #=> sqrt((1 + color(red)(0))/(1+color(red)(0)))# #=> sqrt(1/1)# #= 1# This limit is neither 0 nor infinity, but it's a finite value (#L#). So, by the logic we discussed, #a_n# does the same thing as #b_n#, and since #b_n# diverges, #a_n# also diverges. Hope that helped :)

Christopher Allers · Answer

undefined To apply the limit comparison test for the series $ \sqrt{\frac{n+1}{n^2+2}} $ as $ n $ approaches infinity, follow these steps:

1. Choose a known series that behaves similarly to the given series.
2. Take the limit of the ratio of the given series and the chosen series as $ n $ approaches infinity.
3. Determine the behavior of the limit:
   - If the limit is a positive finite number, the series converge or diverge together.
   - If the limit is 0 or infinity, the series diverge together.

After completing these steps, you'll be able to determine the convergence or divergence of the given series using the limit comparison test.