Question

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

Answer 1

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.

Answer 2

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 :)