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So $1000(2^n+3^n)\le 4^n+5^n$. I take $1000\cdot2^n\le 4^n$ and $1000\cdot3^n\le5^n$ so that adding both gives the inequality, theorem in the ordered fields. So $1000\cdot2^n\le2^2n$. This leads me to an $A$ for $n = 10$ as $1000\cdot2^n < 2^n\cdot2^n$ so $1000\leq2^n$ for $n=10$. But this does not count for $1000\cdot3^n<5^n$. I think to see that $1000\cdot3^n<3^n\left(\frac{5}{3}\right)^n$ so $1000<\left(\frac{5}{3}\right)^n$. so $n = \log_{\frac{5}{3}} (1000)$. This leads me to an $A = 14$ which is my final answer.

Any other suggestions.

A similar problem is $\displaystyle\frac{(1.01)^n}{n^{147}} < 100$. This is tougher. Any hints

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    Kindly read here (http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117) on how to typeset math on this website. It is hard to read what you have written without typeset.2012-05-14
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    The title seems incomplete.2012-05-14
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    I parse the title as "find an $A$ such that for all $n\ge A$ we have $(2^n+3^n)/(4^n+5^n)<1/1000$". The solution makes sense enough. Now, the actual posed question is "a similar problem" about the inequality $1.01^n/n^{147}<100$. The latter inequality is actually *false* for large $n$, so I don't know what the OP is looking for.2012-05-15
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    Sorry about the last inequality it is actually : 1.01^n/N^147 > 100 so > not <. Intuitively it looks strange to me. The denominator seems to increase far more quickly than the nominator. But that is it2012-05-15
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    I have the impression that it actually means <. It could be a typeright error in the handbook. I don't understand what they mean. For me the inequality is true for all n when considered < 100. Every result is far smaller than 1, least to say 100. ????2012-05-15
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    I mentioned it elsewhere. This exercise, and the second mentioned, appear in the handbook, where log, binomials and so on have not been introduced yet. So, it demands for a simple, perhaps although rudimentary solution.2012-07-09

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