Limit $\lim_{n\to +\infty}{\frac{\left(n!\right)^2\cdot4^n\cdot n}{\left(2n\right)!}}$

Question

I am currently trying to calculate the following limit of sequence: $$\lim_{n\to +\infty}{\frac{\left(n!\right)^2\cdot4^n\cdot n}{\left(2n\right)!}}$$ I need it to prove that a series diverges, but I really can't find a way to solve this, any help would be greatly appreciated!

(I really hope I'm not missing anything obvious!)

Edit: You are all absolutely right, I apologize for the lack of context. What is actually going on is that working with a function series, in particular the power series: $$\sum_{n=1}^{\infty}{\frac{\left(n!\right)^2}{\left(2n\right)!}x^n}$$ I needed to find pointwise and total convergence$^1$ for it. To do that I tried to find the radius of convergence using D'Alembert's criterion$^2$ and I found the radius to be $\rho =4$. Therefore my pointwise convergence would be for $\left|x\right|<\rho$.

At this point, I needed to check convergence in the points that coincide with the extremities of the convergence radius, so $x=4$ and $x=-4$. I am currently working on $x=4$. Inserting the value in the original function series I get the following series: $$\sum_{n=1}^{\infty}{\frac{\left(n!\right)^2}{\left(2n\right)!}4^n}$$

I need to find the behaviour of this series. Given how I found the radius of convergence, both the ratio test and the root test are completely useless. I saw Wolfram also uses the integral test, I honestly don't know it, but still, it says it's inconclusive.

I've decided to use the limit comparison test, I don't know if it was the right choice, but either way, since I do know that the series should be divergent, I took $a_n=\frac{\left(n!\right)^2}{\left(2n\right)!}4^n$ and $b_n=\frac{1}{n}$ and that's how I got the limit that this question is about: the limit comparison test will be based on the result of:

$$\lim_{n\to +\infty}{\frac{a_n}{b_n}}=\lim_{n\to +\infty}{\frac{\left(n!\right)^2\cdot4^n\cdot n}{\left(2n\right)!}}=l$$

We know that $\sum_{n=1}^{\infty}{b_n}$ is the harmonic series and is divergent, therefore, if $\lim_{n\to +\infty}{\frac{a_n}{b_n}}=+\infty$ (which, according to Wolfram, is actually the result of the limit) then $\sum_{n=1}^{\infty}{a_n}$ would also be a divergent series. So, yeah, I just need to solve the limit now.

This should be the complete context, if there's something that wasn't clear I don't have any problem clarifying it! And once again, thanks in advance for any response!

Edit 2: In some of the comments there are answers to what I was looking for, but I had already written the context part so I'll post it anyway, as it may be useful to someone someday!

(if you notice I've written anything wrong, please let me know)

$^1$I believe in English it should simply be known as the result of the Weierstrass M-test, in other words, the series converges uniformly and absolutely when there's "total" convergence.

$^2$This one should be the ratio test I think. Since I can't find a lot about it being applied to power series in English resources, I'll specify that given a power series $\sum_{n=0}^{\infty}{a_n x^n}$, the radius of convergence is $\rho =\frac{1}{l}$, where $l$ can be expressed as $l=\lim_{n\to\infty}{\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}}$, when the limit exists.

Answer 1

So, I believe I have come to a working solution, mostly based on the answer provided by @RyszardSzwarc. He deserves the credit, I'll simply be expanding on what he wrote for it to be clearer.

As he said in his answer, we can write $4^n$ as $\left(1+1\right)^{2n}$. We can then use the binomial theorem:

$$\left(x+y\right)^{n}=\sum_{k=0}^{n}{\binom{n}{k}x^{n-k}y^k}\text{, where }\binom{n}{k}=\frac{n!}{k!\left(n-k\right)!}$$

in our specific case, it will become:

$$\left(1+1\right)^{2n}=\sum_{k=1}^{2n}{\binom{2n}{k}}$$

Considering that $0<n<2n$, and that, for this reason, $\binom{2n}{n}$ is one of the terms of our summation, we can get to the following conclusion:

$$\sum_{k=1}^{2n}{\binom{2n}{k}}>\binom{2n}{n}\implies 4^n>\binom{2n}{n}$$

We did all of this for simply one reason:

$$\binom{2n}{n}=\frac{\left(2n\right)!}{n!\left(2n-n\right)!}=\frac{\left(2n\right)!}{\left(n!\right)^2}\implies 4^n>\frac{\left(2n\right)!}{\left(n!\right)^2}$$

At this point, @RyszardSzwarc suggested to simply put the $n$ from the limit into the mix to prove that it does indeed approach $\infty$. This makes sense, but, as @zwim pointed out, the limit goes to infinity without the $n$ too, and this can be proved easily.

If we consider $\frac{\left(n!\right)^2}{\left(2n\right)!}$ for any given $n$, we will end up with a fraction that will have:

• As the numerator, the factorial $n!$
• As the denominator, the factorial $\left(2n\right)!$ divided by $n!$.

If you think about it, dividing $\left(2n\right)!$ for $n!$ is just eliminating the last $n$ terms of $\left(2n\right)!$, for example:

$$n=5\text{, }\quad\frac{\left(n!\right)^2}{\left(2n\right)!}=\frac{5^2\cdot 4^2\cdot 3^2\cdot 2^2\cdot 1^2}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{10\cdot 9\cdot 8\cdot 7\cdot 6}$$

Clearly, when $n\to \infty$, $\frac{\left(n!\right)^2}{\left(2n\right)!}\to 0$. This has some important implications: first of all, it means that the statement: $\lim_{n\to \infty}{\frac{\left(2n\right)!}{\left(n!\right)^2}}=\infty$ is also valid, and second: we have a new way to write our limit:

$$\lim_{n\to \infty}{\frac{\left(n!\right)^2}{\left(2n\right)!}4^n}=\lim_{n\to \infty}{\frac{\left(n!\right)^2/\left(n!\right)^2}{\left(2n\right)!^{}/\left(n!\right)^2}4^n}=\lim_{n\to \infty}{\frac{4^n}{\frac{\left(2n\right)!}{\left(n!\right)^2}}}$$

At this point, all we have to do is to remember the first part of this proof: even tho both parts of the limit are approaching $\infty$, we know that $4^n$ is always greater than $\frac{\left(2n\right)!}{\left(n!\right)^2}$, plus, if we were to calculate the limit of the difference between $4^n$ and $\frac{\left(n!\right)^2}{\left(2n\right)!}$, we would easily find it to be $\infty$.

This means that the two will never converge: they'll only keep accentuating how much faster the first term will approach $\infty$ compared to the second one.

Given all these conditions it appears clear that, as the index $n$ approaches infinity, the limit will inevitably be equal to $\infty$: $4^n$ will be a continually larger number, $\frac{\left(2n\right)!}{\left(n!\right)^2}$ instead, while constantly growing, will also become infinitely small when compared to $4^n$

Edit: I forgot about $n$, clearly when the rest of the limit approaches $\infty$, if we then multiply everything for another sequence that also approaches $\infty$ the limit will still hold true.