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How can I solve this?

I'm trying to prove using limits but it's not working..

Thanks

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    $\begingroup$ have you tried to use the logarithm? in fact above is true because $n$ grows faster than $\log(n)$ $\endgroup$ – user190080 Jul 10 '15 at 13:36
  • $\begingroup$ Yes, I tried to use the logarithm in the ratio limit but i got 1 and not 0. Am I doing something wrong? $\endgroup$ – Leandro Jul 10 '15 at 13:39
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    $\begingroup$ What did you try exactly? We can't tell what you did wrong without seeing what you did. $\endgroup$ – Daniel Fischer Jul 10 '15 at 13:44
  • $\begingroup$ I tried apply logarithm in the ratio= limt n-> inf [log n^a/ log a^n] $\endgroup$ – Leandro Jul 10 '15 at 13:46
  • $\begingroup$ Okay. And can you transform $\dfrac{\log (n^a)}{\log (a^n)}$ into a form that may be easier to handle? $\endgroup$ – Daniel Fischer Jul 10 '15 at 13:49
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so start with for $a>1$ and $\forall n$ big enough we have
$$n^a < a^n \Leftrightarrow \log(n^a)<\log(a^n)\Leftrightarrow a\log(n)<n\log(a)\Leftrightarrow 1<\frac{\log(a)}{a}\frac{n}{\log(n)} $$ and we fixed $a$, so $\frac{\log(a)}{a}=c>0$ is just a constant.

So in fact we have to show, that $$ 1<c\frac{n}{\log(n)} $$ holds for all constants $c>0$ and for all $n\ge n_0(c)$. But this is ofcourse true, since $$ \lim_{n\rightarrow\infty}\frac{n}{\log(n)}=\infty \text{ if we consider }\bar{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,+\infty\} $$ so it grows above each bound.

bests

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Let $x_{n} = a^{n}/n^{a}$ and then we have $$\frac{x_{n + 1}}{x_{n}} = \frac{a^{n + 1}}{(n + 1)^{a}}\cdot\frac{n^{a}}{a^{n}} = a\left(\frac{n}{n + 1}\right)^{a} \to a \text{ as } n \to \infty$$ Now $a > 1$ and hence we can choose a number $k$ with $1 < k < a$ and by the above limit there exists a positive integer $m$ such that $$\frac{x_{n + 1}}{x_{n}} > k$$ for all $n \geq m$. Thus we can see that $$\frac{x_{m + n}}{x_{m}} > k^{n} $$ for all positive integers $n$. Now we can see that $k > 1$ and hence $$k^{n} = (1 + (k - 1))^{n} \geq 1 + n(k - 1) $$ so that $k^{n} \to \infty$ as $n \to \infty$. It follows that $x_{m + n} > x_{m}k^{n}$ and hence $x_{m + n} \to \infty$ as $n \to \infty$. Thus $x_{n} \to \infty$ as $n \to \infty$.

It is now obvious that after a certain value of $n$ we will always have $x_{n} > 1$ and hence $a^{n} > n^{a}$ after a certain value of $n$. There is no need to go for logarithms and complicated limits related to them. Just simple limits coupled with definition of limit is sufficient to handle the problem.

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My attempt at an inductive proof:

Base Case: $n=1 \Rightarrow a^1 > 1$ Which is true by the initial condition

Assumption: $a^k > k^a$

Want to show: $a^{k+1} > (k+1)^a$

Multiply both sides of the assumption by $a$ to get: $a^{k+1} > ak^a$

Thus, it suffices to show that $ak^a > (k+1)^a$

Dividing both sides by the RHS gives: $a(\frac{k}{k+1})^a >1$

For $k$ large enough, the limit of $\frac{k}{k+1}$ is 1, so for sufficiently large $k$ the inequality reduces to $a>1$, which is true

I'm not sure if this is actually correct: can we consider a limit within an inductive proof? If not, is there another way we can finish the proof off?

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