<s>[INST] How would you show that the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$ diverges? How would you show that the series $$\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$$ diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avail: The ratio test is inconclusive, comparing with something "smaller" which I know diverges is proving difficult. If possible, I'd like to show this without using anything fancy (such as Stirling's approximation).
Can you do this with a "basic" comparison?
 [/INST] Here is my attempt:
$a_n=\dfrac{(2n)!}{4^n(n!)^2}$
First i expanded factorial terms:
$a_n=\dfrac{2n(2n-1)(2n-2)\cdots 3\cdot 2\cdot 1}{4^n n(n-1)(n-2)\cdots 2\cdot 1\cdot n(n-1)(n-2)\cdots 2\cdot 1}$
At the numerator, rearrange the $n$ even factors and the $n$ odd factors (not necessary but useful for my description)
$a_n=\dfrac{2n(2n-2)(2n-4)\cdots 4\cdot 2\cdot (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n^2(n-1)^2(n-2)^2\cdots 2^2\cdot 1^2}$
From the  $n$ even factors collect $2$:
$a_n=\dfrac{2^n n(n-1)(n-2)\cdots 2\cdot 1 \cdot (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n^2(n-1)^2(n-2)^2\cdots 2^2\cdot 1^2}$
Now you can simplify the square factors at the denominator with the terms $n(n-1)(n-2)\cdots 2\cdot 1$ obtained from collecting $2$; you have:
$a_n=\dfrac{2^n (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{4^n n(n-1)(n-2)\cdots 2\cdot 1}$
Obviously you can simplify $\dfrac{4^n}{2^n}$; so:
$a_n=\dfrac{(2n-1)(2n-3)(2n-5)\cdots 5\cdot 3\cdot 1}{2^n n(n-1)(n-2)\cdots 2\cdot 1}$
Now consider the factors at the numerator:
$(2n-1)\geq (2n-2)$
$(2n-3)\geq (2n-4)$
$(2n-5)\geq (2n-6)$
and so on.
You have:
$a_n\geq \dfrac{(2n-2)(2n-4)(2n-6)\cdot 4\cdot 2}{2^n n!}$
Again collect a factor $2$ at the numerator:
$a_n\geq\dfrac{2^n (n-1)(n-2)\cdots 4\cdot 2}{2^n n!}$
Simplify the terms $2^n$ and note that the numerator is $(n-1)!$; so:
$a_n\geq\dfrac{(n-1)!}{n!}=\dfrac{1}{n}$
We know that $\sum_{n=1}^\infty\dfrac{1}{n}$ diverges; so by comparison our series:
$\sum_{n=1}^{\infty}a_n$ diverges.
Hope this help.
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