How would you show that the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$ diverges?

Solution 1:

Note ${2n\choose n} >{2^{2n}\over 2n+1}$ because it is the largest binomial coefficient and so is bigger than the average (= sum / total number) and compare to

$$\sum_n{1\over 2n+1}$$

Solution 2:

Note that the terms of your series are

$$ \frac{1}{2}, \quad \frac{1}{2} \times \frac{3}{4}, \quad \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6}, \quad \cdots $$

Compare this with the sequence

$$ \frac{1}{2}, \quad \frac{1}{3}, \quad \frac{1}{4}, \quad \cdots $$

which can be rewritten

$$ \frac{1}{2}, \quad \frac{1}{2} \times \frac{2}{3}, \quad \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}, \quad \cdots $$

Since the first terms are equal, and the ratios between successive terms are always larger in the first series than in the second, it follows that the sum of the first series (if it exists) must be larger than the sum of the second. Since the second diverges (it is simply the harmonic series minus the first term), so must the first.