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

Apr 13, 2018

Diverges when compared to ${b}_{n} = \frac{1}{\sqrt{n}}$

#### Explanation:

We need to come up with a new sequence ${b}_{n}$ to compare to ${a}_{n} = \sqrt{\frac{n + 1}{{n}^{2} + 2}}$. Furthermore, we need to be able to easily determine the divergence of ${\sum}_{n = 1}^{\infty} {b}_{n}$. 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.

${b}_{n} = \sqrt{\frac{n}{n} ^ 2} = \sqrt{\frac{1}{n}} = \frac{1}{\sqrt{n}} = \frac{1}{n} ^ \left(\frac{1}{2}\right)$

Now, we know ${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{1}{2}\right)$ diverges by the $p -$series test, $p = \frac{1}{2} < 1$.

The Limit Comparison tells us if we know the convergence or divergence of ${a}_{n}$ or ${b}_{n}$ and

$c = {\lim}_{n \to \infty} {a}_{n} / {b}_{n} > 0 \ne \pm \infty$, then both series either converge or diverge.

Knowing ${a}_{n} = \sqrt{\frac{n + 1}{{n}^{2} + 2}} , {b}_{n} = \frac{1}{\sqrt{n}} , {\sum}_{n = 1}^{\infty} {b}_{n}$ diverges,

$c = {\lim}_{n \to \infty} \frac{\sqrt{\frac{n + 1}{{n}^{2} + 2}}}{\frac{1}{\sqrt{n}}} = {\lim}_{n \to \infty} \sqrt{\frac{n \left(n + 1\right)}{{n}^{2} + 2}} = {\lim}_{n \to \infty} \sqrt{\frac{{n}^{2} + n}{{n}^{2} + 2}} = 1 > 0 \ne \pm \infty$

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:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n}$

${a}_{n} \ge 0 , {b}_{n} > 0$ for all $n$.

If this limit $\textcolor{red}{= 0}$, then $\textcolor{red}{\text{both series converge}}$
If this limit $\textcolor{g r e e n}{= L}$ (any arbitrary nonzero value), then $\textcolor{g r e e n}{{a}_{n} \text{ does what " b_n " does.}}$
If this limit $\textcolor{b l u e}{\text{goes to infinity}}$, then $\textcolor{b l u e}{\text{both series diverge}}$.

So now, we figure out what series it would be ideal to compare this to. I'm going to chose $\sqrt{\frac{n}{n} ^ 2}$ which, with some simplification, turns into $\frac{1}{\sqrt{n}}$. By the p-test, we know that this series diverges. Bearing this in mind, let's take the limit:

$\implies {\lim}_{n \to \infty} \frac{\sqrt{\frac{n + 1}{{n}^{2} + 2}}}{\frac{1}{\sqrt{n}}}$

$\implies {\lim}_{n \to \infty} \sqrt{\frac{{n}^{2} + n}{{n}^{2} + 2}}$

Now, we just evaluate this limit using the same steps we learned in Calc 1. We just divide every term by the highest power:

$\implies {\lim}_{n \to \infty} \sqrt{\frac{{n}^{2} / {n}^{2} + \frac{n}{n} ^ 2}{{n}^{2} / {n}^{2} + \frac{2}{n} ^ 2}}$

$\implies {\lim}_{n \to \infty} \sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n} ^ 2}}$

..and now take the limit as $n \to \infty$ of each term

$\implies \sqrt{\frac{1 + \textcolor{red}{0}}{1 + \textcolor{red}{0}}}$

$\implies \sqrt{\frac{1}{1}}$

$= 1$

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 :)