# How to solve the limit $\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$.

How to solve this limit??

$$\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$$

It's a limit, not a series

\begin{align} \lim_{k\to\infty} \frac{(2k)!}{2^{2k}\cdot(k!)^2} &=\lim_{k\to\infty} \frac{(2k)!}{2^k \cdot 2^k \cdot k! \cdot k!} \\ &=\lim_{k\to\infty} \frac{(2k)!}{(2^k \cdot k!)^2} \\ &=\lim_{k\to\infty} \frac{(2k)(2k-1)\cdots(2)(1)}{(2k)^2 (2k-2)^2 \cdots (4)^2 (2)^2} \\ &=\lim_{k\to\infty} \frac{(2k-1)(2k-3)\cdots(1)}{(2k)(2k-2)\cdots(2)} \\ &=0 \end{align}
In the last step, you can think of the fraction as the infinite product of fractions less than $$1$$ ($$\frac{1}{2} \times\frac{3}{4} \times \frac{5}{6} \times\frac{7}{8}...$$), which will decrease to $$0$$.
• Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here. – coffeemath Nov 22 '18 at 7:16
This is the probability of having $$n$$ heads and $$n$$ tails in $$2n$$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.