Is the first function big oh, theta or omega of the second function in the following examples.

$(\log n)^{\log n} $ and $ \frac{n}{\log n}$
I am not sure how fast the first function grows.

$n2^n$ and $3^n$
I think the second function grows faster but not sure how to prove it.

$(\log n)^{\log n} $ and $2^{(\log_2n)^2}$
This is just puzzling.

$\sum_{i=1 }^n i^k$ and $n^{k+1}$.
And so is this.

Any help is much appreciated.


1 Answer 1


For the first and the third, taking the logarithm of both sides will really help. For the second, try to use the fact that if $\lim_{n\to\infty} \frac{f(n)}{g(n)} = 0$ then $f(n) = o(g(n))$. For the fourth, try approximating the sum using an integral.

  • $\begingroup$ for the first example after taking log of both sides we get $ lognlog(logn) $ and $logn - log(logn) $. Do we say that $log(logn)$ is approximately equal to 1, therefore the two functions are asymptotically equivalent? Thanks for your help $\endgroup$
    – matt
    Commented Apr 10, 2017 at 9:58
  • 2
    $\begingroup$ The function $\log\log n$ is definitely not approximately equal to 1. I suggest working it through with a friend or a TA. Also, don't forget that you took the logarithm – even if $\log f = \Theta(\log g)$ it doesn't mean that $f = \Theta(g)$. For example, $\log(2^n) = \Theta(\log(3^n))$. $\endgroup$ Commented Apr 10, 2017 at 10:29

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