Wolfram tells me that the the limit is $0$ when $n$ goes to infinity. Unfortunately, I have no idea how to prove it...
$\lim_{n\to\infty}\frac {2^\sqrt { \log(\log n)}}{\log n}.$
Any help would be appreciated, thanks in advance.
Wolfram tells me that the the limit is $0$ when $n$ goes to infinity. Unfortunately, I have no idea how to prove it...
$\lim_{n\to\infty}\frac {2^\sqrt { \log(\log n)}}{\log n}.$
Any help would be appreciated, thanks in advance.
Hints:
The logarithm of this quantity is $\log 2\cdot\sqrt{\log(\log n)}-\log(\log n)$.
When $n\to+\infty$, $\log(\log n)\longrightarrow$ $_________$.
When $x\to+\infty$, $\log2\cdot\sqrt{x}-x\longrightarrow$ $_________$.
Hence $\log2\cdot\sqrt{\log(\log n)}-\log(\log n)\longrightarrow$ $________$ when $n\to+\infty$.
And finally $2^{\sqrt{\log(\log n)}}/\log n\longrightarrow$ $_________$ when $n\to+\infty$.
Starting with $f(n) = \frac{2^{\sqrt{\log(\log(n))}}}{\log(n)}$ and taking logarithms $\log(f(n)) = \log{2}{\sqrt{\log(\log(n))}} - \log({\log(n)})=\sqrt{\log(\log(n))}\left(\log(2)-\sqrt{\log(\log(n))}\right)$
then $\sqrt{\log(\log(n))}$ increases towards $+\infty$ with increasing $n$, while $\left(\log(2)-\sqrt{\log(\log(n))}\right)$ heads towards $-\infty$, so $\log(f(n))$ heads towards $-\infty$ and $f(n)$ heads towards $0$.