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This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$. Solve for n explicitly without calculator: $$\frac{3^n}{n!}\le10^{-6}$$ And I appreciate hint rather than explicit solution. Thank You. |
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Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] <\frac{2}{9}\left(m!\right)^3.$$ So you have to go at least far enough so that $$ \frac{2}{9}\left(m!\right)^3>10^{6}, $$ or $m! > \sqrt[3]{4500000} > 150$. So $m=5$ (corresponding to $n=15$) isn't far enough; the smallest $n$ satisfying your inequality will be at least $16$. Similarly, for $n=3m+1$, $$ 3^{-3m-1}(3m+1)!=\left[\left(m+\frac{1}{3}\right)m\left(m-\frac{1}{3}\right)\right]\cdots \left[\frac{4}{3}\cdot1\cdot\frac{2}{3}\right]\cdot\frac{1}{3} < \frac{1}{3}(m!)^3,$$ so you need $m!>\sqrt[3]{3000000}> 140$, and $m=5$ (that is, $n=16$) is still too small. Finally, for $n=3m+2$, $$ 3^{-3m-2}(3m+2)!=\left[\left(m+\frac{2}{3}\right)\left(m+\frac{1}{3}\right)m\right]\cdots \left[\frac{5}{3}\cdot\frac{4}{3}\cdot1\right]\cdot\frac{2}{3}\cdot\frac{1}{3} > \frac{560}{729}(m!)^3, $$ where the coefficient comes from the last eight terms, so it is sufficient that $m! > 100\cdot\sqrt[3]{729/560}.$ To show that $m=5$ is large enough, we need to verify that $(12/10)^3=216/125 > 729/560$. Carrying out the cross-multiplication, you can check without a calculator that $216\cdot 560 =120960$ is larger than $729\cdot 125=91125$, and conclude that $m=5$ (that is, $n=17$) is large enough. The inequality therefore holds for exactly all $n\ge 17$. |
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I would use Stirling's approximation $n!\approx \frac {n^n}{e^n}\sqrt{2 \pi n}$ to get $\left( \frac {3e}n\right)^n \sqrt{2 \pi n} \lt 10^{-6}$. Then for a first cut, ignore the square root part an set $3e \approx 8$ so we have $\left( \frac 8n \right)^n \lt 10^{-6}$. Now take the base $10$ log asnd get $n(\log 8 - \log n) \lt -6$ Knowing that $\log 2 \approx 0.3$, it looks like $16$ will not quite work, as this will become $16(-0.3)\lt 6$. Each increment of $n$ lowers it by a factor $5$ around here, or a log of $0.7$. We need a couple of those, so I would look for $18$. Added: the square root I ignored is worth a factor of $10$, which is what makes $17$ good enough.a Alpha shows that $17$ is good enough. |
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How about we overestimate $3^n$ as $\sqrt{10}^n$ and underestimate every contribution to the factorial beyond $10$ as only $10$? Then $$\frac{3^n}{n!} \le \frac{10^5 \cdot 10^{\frac{n-10}2}}{10! \cdot 10^{n-10}} $$ Since $10!$ is about $10^{6.5}$ we have $$ 10^{-1.5} \cdot 10^{-\frac{n-10}2} \le 10^{-6} $$ After taking the $\log_{10}$ we have $$ -1.5 -\frac{n-10}2 \le -6$$ $$ 1.5 + \frac{n-10}2 \ge 6$$ $$ \frac{n-10}2 \ge 4.5$$ $$ n-10 \ge 9 $$ $$ n \ge 19$$ Since you asked for a solution and not the minimal solution this should suffice, and is more plausibly done without a calculator. The only real mental calculation was remembering the approximate value of $10!$ (well, and $3^2$, but let's be reasonable). Note that if you know the factorials for higher numbers (or exact values for some powers of 3, such as $3^7 = 2187$) then you could refine this argument using higher $n$ for the "fixed part". |
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how about make a function?$f(n)=\frac{3^n}{n!}-10^{-6}$ or maybe $f(n)=\frac{3^n}{n!10^{-6}}$ |
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