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I think it should be like:

$=\dfrac{\log(\log(n^2+10))}{\log(n)}$

$=\dfrac{\log(2\log(n))}{\log(n)}$

$=\dfrac{\log(\log(n))}{\log(n)}$

But $\log(\log(n))<\log(n)$ As the $log(n)$ is in the denominator.

Hence it is $O(\log(\log(n)))$

Is it correct?

David
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  • In fact, and more importantly, $\frac{\log(\log(n^2+1))}{\log(n)}$ is $O(1)$, in other words there exists a $C>0$ so that for all $n$ large enough, $$\frac{\log(\log(n^2+1))}{\log(n)} < C(1)$$

    The fact it is $O(1)$ implies that it is $O(\log(\log(n))$ because $O(1) \subsetneq O(\log(\log(n))$,

     – adfriedman Dec 13 '17 at 18:03
  • The fact that this holds for any $C>0$ means that it is also $o(1)$. – adfriedman Dec 13 '17 at 18:12

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