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I want to prove that the following sequence is monotonously decreasing:

$A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$

I think it should be pretty easy to do so by induction:

  • $A_1 = \binom{3}{2}\cdot{(\frac{60}{100})}^2\cdot{(\frac{40}{100})}^1+\binom{3}{3}\cdot{(\frac{60}{100})}^3\cdot{(\frac{40}{100})}^0 = 0.648$

  • $A_2 = \binom{6}{4}\cdot{(\frac{60}{100})}^4\cdot{(\frac{40}{100})}^2+\binom{6}{5}\cdot{(\frac{60}{100})}^5\cdot{(\frac{40}{100})}^1+\binom{6}{6}\cdot{(\frac{60}{100})}^6\cdot{(\frac{40}{100})}^0 = 0.54432$

  • Let's assume that $A_{k} > A_{k-1}$

  • Let's prove that $A_{k+1} > A_{k}$

So how do I go about with that? Or is there another way?

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This isn't rigorous, and it may in fact be what motivates the question, but you can interpret the $A_k$'s as the probability that a biased coin that lands Heads $60\%$ of the time and Tails $40\%$ of the time will come up Heads at least two thirds of the time when tossed $3k$ times. Since $60\%$ is less than $2/3=66.666...\%$, you can expect this to become less and less likely as $k$ gets larger.

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  • $\begingroup$ It is indeed what motivated the question. In fact, it is my answer math.stackexchange.com/a/870044/131263 to someone else's question... Thanks (BTW, you might be able to see a different way to answer that other question). $\endgroup$ – barak manos Jul 17 '14 at 16:44
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It happens that $$ A_k = \left(\frac{18}{125}\right)^k\binom{3k}{2k}\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right),$$ and a clever idea is to use the Stirling approximation together with the Gauss continued fraction for the hypergeometric function in order to give tight bounds for $A_k$, then prove $A_{k+1}<A_{k}$ for any $k$ big enough. $$\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right)=1+\frac{\frac{3k}{2+4k}}{1-\frac{\frac{3k-3}{4+4k}}{1+\frac{3k-3}{4+4k}+\frac{\frac{3(2-k)}{2(3+2k)}}{1+\ldots}}}\in\left[4-\frac{60}{k+20},4-\frac{60}{k+53}\right],$$ so the asymptotic behaviour of $A_k$ is: $$A_k=\left(\frac{243}{250}\right)^k\sqrt{\frac{12}{\pi k}}\left(1+\Theta\left(\frac{1}{k}\right)\right)$$ and $\{A_k\}_{k\in\mathbb{N}}$ decreases for sure from a certain point on. If we compute the constants hidden in the $\Theta$-notation, we can also notice that $$\left(\frac{250}{243}\right)^k A_k$$ is a decreasing function, hence $A_k$ is a decreasing function, too.

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