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Does $n^{1000000}/2^{n}$ converge as $n\rightarrow+\infty$?

To solve this problem, I am only allowed to use elementary properties of limits. This is what I tried. For $n > 1000000$, one has that \begin{align*} \frac{n^{1000000}}{2^{n}} = \left(\frac{n}{2}\right)^{1000000}\times\left(\frac{1}{2}\right)^{n-1000000} \end{align*}

But then I get stuck. This is not homework. Could someone help me out?

user0102
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    hava a look at https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial – ThomasL Mar 02 '20 at 22:20
  • A general rule of thumb is that exponential functions "eventually grow faster" than geometric functions. So the limit is $0$. – fleablood Mar 02 '20 at 22:33
  • Dear @Maximilian Janisch, thanks for the reference. But I think the given answer is a little bit overkill. The exercise whence I got this problem is at the very beginning of the theory of real-valued sequences. – user0102 Mar 02 '20 at 22:42
  • @BrickByBrick I see what you are saying but I think that you can find a useful and elementary answer to your question there – Maximilian Janisch Mar 02 '20 at 22:43

2 Answers2

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Taking logarithms of the terms gives $100000\ln n-n\ln 2$. This tends to $-\infty$ since $$\frac{\ln n}{n}\rightarrow 0$$ and so the terms are tending to $0$.

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Hint: Think about relative growth rates of something simpler like $n^{10}$ versus $2^{n}$ and work your way up.

Donja362
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