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

2 Answers
Apr 13, 2018

Diverges when compared to b_n=1/sqrtn

Explanation:

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.

Apr 13, 2018

The series diverges.

See work below:

Explanation:

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 :)