How to solve this limit??
$$\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}$$
It's a limit, not a series
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I found that the answer is 0, however, I need to prove it. – S. Yoo Nov 22 '18 at 05:26
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1HINT: note that $2^{2k}(k!)^2=\left(\prod_{j=1}^k 2j\right)^2=(2n(2n-2)\cdots 2)^2$ – Nov 22 '18 at 05:30
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https://en.wikipedia.org/wiki/Stirling%27s_approximation – lab bhattacharjee Nov 22 '18 at 05:35
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The SAME limit is here: https://math.stackexchange.com/questions/788096/how-to-prove-that-lim-n-to-infty-frac2n-12n-0 – Robert Z Nov 22 '18 at 05:38
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Possible duplicate of How to prove that $\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0$ – Nosrati Nov 22 '18 at 05:48
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\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 $0$.
Rócherz
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3Not 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 07:16
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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.
Mostafa Ayaz
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