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?

The sum $\sum_{n=1}^\infty\frac{(2n)!}{4^n(n!)^2}=\sum_{n=1}^\infty\binom{2n}n(\frac12)^{2n}$ is the expected number of returns to the origin in a symmetric random walk on the integers. [*] Instead of trying to show analytically that the series diverges, let's try to use direct probabilistic reasoning to see why the expectation is infinite. Equivalently, we wish to show that, at any point, the probability of returning to the origin is one.
Let $a_0=1$ and, for $n\gt0,$ let $a_n$ be the probability of returning to the origin from a point at a distance of $n$ units from the origin. Since the walk is symmetric, for $n\gt0$ we have $a_n=\frac12a_{n-1}+\frac12a_{n+1},$ i.e., $a_{n-1},a_n,a_{n+1}$ are in arithmetic progression. Hence the sequence $a_0,a_1,a_2,\dots$ is an infinite arithmetic progression. Since the terms $a_n$ are probabilities, the progression is confined to the interval $[0,1],$ so it can only be a constant sequence. Since $a_0=1,$ we must have $a_n=1$ for all $n,$ Q.E.D.
[*] Consider an infinite sequences of independent fair coin tosses. Let the indicator variable $X_n$ take the value $1$ if we have equal numbers of heads and tails in the first $2n$ tosses, $0$ otherwise. Then the number of returns to the origin is the random variable $X=\sum_{n=0}^\infty X_n,$ and by linearity $E(X)=\sum_{n=1}^\infty E(X_n)=\sum_{n=1}^\infty P(X_n=1)=\sum_{n=1}^\infty\binom{2n}n(\frac12)^{2n}.$
