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How can I solve the flowing limit?

$$\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=0$$

Attempt 1:

$\log_2 n = m \implies \log_2 n^2 = 2\cdot\log_2 n = 2\cdot m$ $$\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=\lim\limits_{n \to \infty} \frac{n \log_2(m)}{3^{2 \cdot m}}$$

$\log_2(m) = r \iff 2^r = m$

$$\lim\limits_{n \to \infty} \frac{n \log_2(m)}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{n\cdot 2^r}{3^{2 \cdot m}}$$

$m = \log_2n \iff2^m=n$

$$\lim\limits_{n \to \infty} \frac{n\cdot 2^r}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{2^m\cdot 2^r}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{2^{r+m}}{9^{m}} =0$$
I'm not sure if it is so right.
Are there other ways of solve this limit? Thanks!

Attempt 2 (By L'Hopital's rule):

Let $f(n) = 3^{\log_2 n^2}$ and $g(n) = n \log_2( \log_2 n) \implies \lim\limits_{n \to \infty} f(n) = \infty \land \lim\limits_{n \to \infty} g(n) = \infty$

$$\lim\limits_{n \to \infty} \frac{g(n)}{f(n)}\implies \lim\limits_{n \to \infty} \frac{g´(n)}{f´(n)}$$

I get:

$f´(n)= \frac{dn}{n}(3^{\log_2 n^2}) = \frac{ 3^{\log_2 n^2} 2 \log(3)}{n \log(2)}$

$g´(n)= \frac{dn}{n}(n \log_2( \log_2 n)) = \frac{\log(n)\log(\log_2n)+1}{\log(2) \log(n)}$

also

$$\lim\limits_{n \to \infty} \frac{g´(n)}{f´(n)} = \lim\limits_{n \to \infty} \frac{\frac{\log(n)\log(\log_2n)+1}{\log(2) \log(n)}}{\frac{ 3^{\log_2 n^2} 2 \log(3)}{n \log(2)}} = \lim\limits_{n \to \infty} \frac{(\log(n)\log(\log_2n)+1)\cdot(n \log(2))}{(\log(2) \log(n))\cdot(3^{\log_2 n^2} 2 \log(3))}$$

This does not help me :(
Which is the most optimal form to solve this limit?

Darío A. Gutiérrez
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    Solve? Do you mean you are supposed to show that the limit is 0? Since the top and bottom both seem to go to infinity, can't you use l'hopital's rule (sp?). – Arby Mar 21 '17 at 14:02
  • In your last limit your expression contains only $m$ and $r$ but the limit is still for $n$ going to infinity...Limit does go to zero though – imranfat Mar 21 '17 at 14:17
  • Does this help? $\lim\limits_{n \to \infty} \frac{2^m\cdot \log_2 m}{3^{2 \cdot m}}=\lim\limits_{n \to \infty} \frac {2^m\cdot \log_2 m} {3^m\cdot3^m}=\lim\limits_{n \to \infty} \frac {2^m} {3^m}\cdot\frac {\log_2 m} {3^m}$ Note that if $n$ goes to infinity that $m=\log_2 n$ does as well. – Arby Mar 21 '17 at 14:19
  • Of course the only hope to transform this into a proof is to justify the last step, that is, to explain why $$\lim_{n\to\infty}2^{\log_2\log_2n+\log_2n}9^{-\log_2n}=0$$ Any idea to prove that? – Did Mar 21 '17 at 14:20
  • Your first substitution is correct, you just have to get through with it. You didn't substitute $n = 2^m$ in the RHS. The limit is then as $m \to +\infty$. – rubik Mar 21 '17 at 14:22
  • Many thanks for their answer, but I can not find the optimal solution and the L' Hopital's rule makes it more difficult. – Darío A. Gutiérrez Mar 21 '17 at 21:58

2 Answers2

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$\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=0 $

We have

$\begin{array}\\ 3^{\log_2 n^2} &=e^{\ln 3\log_2 n^2}\\ &=e^{2\ln 3\log_2 n}\\ &=e^{2\ln 3 \ln(n)/\ln 2}\\ &=e^{c \ln(n)} \qquad\text{where }c = 2\ln 3/\ln 2 > 1\\ &=(e^{ \ln(n)})^c\\ &= n^c\\ \end{array} $

so $\frac{n \log_2( \log_2 n)}{n^c} =\frac{\log_2( \log_2 n)}{n^{c-1}} \to 0 $ since $\frac{\ln(n)}{n^a} \to 0$ for any $a > 0$.

Darío A. Gutiérrez
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marty cohen
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Let $u=\log_{2}(n)$ and notice that $n=2^u$. $$\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{\log_{2}(n^2)}}=\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{2\log_{2}(n)}}=$$ $$\lim_{u\to\infty}\frac{2^u\log_{2}(u)}{3^{2u}}=\lim_{u\to\infty}\frac{2^u\log_{2}(u)}{9^u}=$$ $$\lim_{u\to\infty}(\frac{2}{9})^u\log_{2}(u)=\lim_{u\to\infty}\frac{\ln(u)}{\ln(2)(\frac{9}{2})^u}$$

Now we get an indeterminate form of $\frac{\infty}{\infty}$, so we can apply L'Hospital's Rule.

$$\lim_{u\to\infty}\frac{\ln(u)}{\ln(2)(\frac{9}{2})^u}=\lim_{u\to\infty}\frac{\frac{d}{du}(\ln(u))}{\ln(2)(\frac{d}{du}(\frac{9}{2})^u)}=$$ $$\lim_{u\to\infty}\frac{\frac{1}{u}}{\ln(2)\ln(\frac{9}{2})(\frac{9}{2})^u}=\lim_{u\to\infty}\frac{1}{\ln(2)\ln(\frac{9}{2})(\frac{9}{2})^uu}=0$$

So, $$\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{\log_{2}(n^2)}}=0$$

Jonathan Barkey
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