Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded.

Question

I'm trying to prove the following:

Let $n\in \mathbb N$ and $$ x_n = \frac{(2n)!!}{(2n-1)!!} $$ Where: $$ (2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\ (2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) $$ Prove $\{x_n\}$ is not bounded.

Intuitively it feels like it is unbounded. Let's consider the following fraction:

$$ x_n = \frac{2\cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) } $$

Now if we consequently take a number from nominator and denominator we get:

$$ \frac{2}{1} > 1 \\ \frac{4}{3} > 1 \\ \frac{6}{5} > 1 \\ \dots \\ \frac{2n}{2n-1} > 1 $$

So the fraction is a product of rational numbers each of which is greater than $1$ and the product of rational numbers greater than $1$ is increasing.

I've tried to formalize that by expanding $(2n)!!$ and $(2n-1)!!$:

$$ (2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2) = 2^kn!\\ (2n-1)!! = (2n-1)(2n-3)\cdots(5)(3)(1) = \\ = \frac{(2n-1)(2n-2)(2n-3)\cdots(5)(4)(3)(2)(1)}{(2n-2)(2n-4)\dots(4)(2)} = \\ \frac{(2n-1)!}{2^{n-1}(n-1)!} $$

So using the above:

$$ x_n = \frac{2^nn!2^{n-1}(n-1)!}{(2n-1)!} = \frac{2^{2n-1}n!(n-1!)}{(2n-1)!} $$

Here is where I got stuck. How do i proceed with the proof using some constant $M$ and some number $N > n$ such that $x_N > M$? Should i introduce some inequality?

I've seen a similar question, but the sequence there is proven to be divergent which i guess is more a calculus concept (yet very similar to (un)boundedness), and i'm in search of a precalculus solution.

Answer 1

As simple as it gets:

LEMMA: For $k\in N$ and $k>1$

$$\frac{2k}{2k-1}\gt\frac{\sqrt{3k+1}}{\sqrt{3k-2}}\tag{1}$$

Proof: It's trivial. Both sides are positive, square them:

$$\frac{4k^2}{4k^2-4k+1}\gt\frac{3k+1}{3k-2}$$

..which is equivalent to:

$$4k^2(3k-2)\gt(4k^2-4k+1)(3k+1)$$

$$12k^3-8k^2\gt12k^3+4k^2-12k^2-4k+3k+1$$

$$12k^3-8k^2\gt12k^3-8k^2-k+1$$

$$0\gt-k+1$$

...which is obviously true.

End of lemma proof.

By using the inequality (1) we have:

$$\frac{2}{1}=\frac{\sqrt{4}}{\sqrt{1}}$$

$$\frac{4}{3}\gt\frac{\sqrt{7}}{\sqrt{4}}$$

$$\frac{6}{5}\gt\frac{\sqrt{10}}{\sqrt{7}}$$

$$...$$

$$\frac{2n-2}{2n-3}\gt\frac{\sqrt{3n-2}}{\sqrt{3n-5}}\tag{1}$$

$$\frac{2n}{2n-1}\gt\frac{\sqrt{3n+1}}{\sqrt{3n-2}}\tag{1}$$

Multiply all this and you get:

$$a_n=\frac{(2n)!!}{(2n-1)!!} > \sqrt{3n+1}$$

So $a_n$ is obviously unbounded. Cute, isn't it? :)

Answer 2

This answer uses calculus-based language. It is likely not quite what the OP was looking for (requesting a pre-calculus solution), but I'm posting so that someone else will suggest a way to avoid the use of l'Hopital's rule.

Observe that $x_n$ is unbounded if and only if $\ln(x_n)$ is unbounded. Since $x_n$ is a product, we can write $$ \ln(x_n)=\sum_{k=1}^{n}\ln\left(\frac{2k}{2k-1}\right). $$

If we show that $$ \sum_{k=1}^{\infty}\ln\left(\frac{2k}{2k-1}\right) $$ exists (i.e., converges), then the $\ln(x_n)$'s must be bounded as they are the partial sums of this sum. We will see, however, that the opposite is true.

We will use the limit comparison test and compare this series to the (divergent) harmonic series with entries $\frac{1}{k}$. Therefore, we compute $$ \lim_{k\rightarrow\infty}\frac{\ln\left(\frac{2k}{2k-1}\right)}{\frac{1}{k}}. $$ This limit is an indeterminate form $\frac{0}{0}$, so we can use l'Hopital's rule (the only Calculus part of this answer) to get that this limit equals $$ \lim_{k\rightarrow\infty}\frac{\frac{2}{2k}-\frac{2}{2k-1}}{-\frac{1}{k^2}}= \lim_{k\rightarrow\infty}\frac{\frac{-2}{2k(2k-1)}}{-\frac{1}{k^2}}= \lim_{k\rightarrow\infty}\frac{2k^2}{2k(2k-1)}=1. $$ Therefore, both series diverge by the limit comparison test. Since $\sum\frac{1}{k}$ diverges to infinity, so does $\sum\ln\left(\frac{2k}{2k-1}\right)$, i.e., $\ln(x_n)$ is unbounded.

Answer 3

If you take

$$\log(1-x)\le-x$$ for granted*,

$$\log x_n=\sum_{k=1}^n\log\left(\frac{2k}{2k-1}\right)=-\sum_{k=1}^n\log\left(1-\frac1{2k}\right)>\frac12\sum_{k=1}^n\frac1k$$

which diverges.

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You can draw it from

$$\log(1+0)\le-0$$ and

$$(\log(1+x))'=\frac1{1+x}\le(-x)'=-1.$$

Answer 4

Without integrals or derivatives: $$ x_n=\frac{(2n)!!}{(2n-1)!!}= \prod_{k=0}^{n-1}\left(1+\frac{1}{2k+1}\right)=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$ implies $$ x_n^2 = \prod_{k=1}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)^{-1}=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)^{-1}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right) $$ or $$ x_n^2 = \frac{n}{4}\prod_{k=1}^{n-1}\left(1-\frac{1}{(2k+1)^2}\right). $$ On the other hand the infinite product $\prod_{k\geq 1}\left(1-\frac{1}{(2k+1)^2}\right)$ is convergent to a positive constant since $\sum_{k\geq 1}\frac{1}{(2k+1)^2}$ is convergent to a positive constant, hence $x_n\geq \sqrt{K n}$ for some mysterious constant $K>0$, which actually equals $\pi$.