How do you prove that this is divergent?

My lecturer has challenged me to complete this question. I need to prove the series either diverges or converges. I know that the series is divergent by using divergent tests, but I need to do this by the comparison test. The problem is the -3 in the numerator and the only hint given is that I need to divide the numerator by something to make n^4 smaller.

Anybody know how to do this?

So you just need a lower bound on $\displaystyle \left( \dfrac{n^4-3}{n^6 + n - 1} \right)^{1/2}$ whose series diverges.

Just divide numerator and denominator by $n^4$ to obtain


$\displaystyle \left( \dfrac{1-3n^{-4}}{n^2 + n^{-3} - n^{-4}} \right)^{1/2}$

Note that $1-3n^{-4} \geq c_1$ for $n \geq 2$

and $n^2 + n^{-3} - n^{-4} < c_2 n^2$

where I leave it to you to determine what $c_1,c_2$ are as they are constants.


Hence bound is $\displaystyle \sum_{n=2}^{\infty} \left( \dfrac{n^4-3}{n^6 + n - 1} \right)^{1/2} > \sqrt{ \dfrac{c_1}{c_2} } \sum_{n=2}^{\infty} \dfrac{1}{n}$

and clearly the lower bound diverges.

Thankyou for your reply! Based off what you said, would something like this work?

You are working backward from what I said even though I intended for you to go *from* the LHS *to* the RHS.

So $1-3n^{-4} \geq 1-3 \cdot 2^{-4} = \dfrac{13}{16} = c_1$

and similarly proceed to show that

$n^2 + n^{-3} - n^{-4}$ has an upper bound of $c_2 n^2$ for some $c_2$.

P.S. And no, the bottom half of your page isn't valid so don't bother substituting in n=2 for this one.

Oh okay, so for c2 something like this

That's not a convincing argument and not what I was suggesting. While that value of $c_2$ works, you need to refine your argument.


For all $n \geq 2$ we have that

$1 < n^2 + n^{-3} - n^{-4} < n^2 + n^{-3} < n^2 + kn^2$

and really, you could just pick $k = 1$ at this stage (because it's obvious that $n^{-3} < n^2$) hence obtain $c_2 = 2$.

We are effectively bounding $n^{-3} < kn^2$ so it's just a matter of having $k > n^{-5}$ for all $n \geq 2$. So just choose $k > \displaystyle \max_{n \geq 2} n^{-5} = 2^{-5}$ hence the bound $n^2 + n^{-3} < c_2 n^2$ is guaranteed if we just choose

$c_2 = 1+k > \dfrac{33}{32}$.

Your answer you got through a bad argument works because $\dfrac{65}{4} > \dfrac{33}{32}$.

Hi, sorry I'm struggling to understand this as it's not usually how we complete the questions. If you have time, could you try and explain this is as much detail as possible?

In order to find the lower bound for $\left( \dfrac{1-3n^{-4}}{n^2 + n^{-3} - n^{-4}} \right)^{1/2}$ you need to do two things;

(a) find a lower bound for $1-3n^{-4}$

(b) find an upper bound for $n^2 + n^{-3} - n^{-4}$.


For (a), it is obvious that $1-3n^{-4}$ is an increasing function for $n \geq 2$ therefore we are justified to sub in $n=2$ to obtain the lower bound. Hence $1-3n^{-4} \geq \dfrac{13}{16}$.

For (b), it is less so obvious as to how this function behaves. We cannot immediately say that this is a decreasing function, so we cannot substitute anything in. But we can notice by inspection that $n^{-4}$ is being taken away, therefore $n^2 + n^{-3}$ is a bigger number.
Hence $n^2 + n^{-3} - n^{-4} < n^2 + n^{-3}$.
Now also notice that for $n \geq 2$ we have that $n^{-3}$ is a very small number. We are adding it onto $n^2$. But we add on $n^2$ instead, then this is obviously going to give as a bigger result because $n^2$ is much bigger than $n^{-4}$. Therefore;
$n^2 + n^{-3} - n^{-4} < n^2 + n^{-3} < n^2 + n^2 = 2n^2$.

Hence the upper bound here is $2n^2$. And so the overall bound becomes; $\left( \dfrac{1-3n^{-4}}{n^2 + n^{-3} - n^{-4}} \right)^{1/2} > \left( \dfrac{13}{32n^2} \right)^{1/2} = \sqrt{\dfrac{13}{32}} \cdot \dfrac{1}{n}$

and its series is a divergent one, hence by comparison test $\displaystyle \sum_{n \geq 2} \left( \dfrac{n^4-3}{n^6 + n - 1} \right)^{1/2}$ diverges.




I merely introduced $k$ to show you that we can bound $n^{-3}$ by *not just* $n^2$ but by any $kn^2$ as long as $k > \dfrac{1}{32}$, and since your method was incorrect, it still obtained a valid coefficient because this inequality was satisfied by it.