How is $n^{1.001} + n\log n = \Theta (n^{1.001})$? I am studying for an exam and stumbled across this here:
https://cs.stackexchange.com/questions/13451/few-big-o-example
(I cant comment there since commenting needs 50 reps and I am a new user. Thought math exchange would help)
The chosen answer says for large $n$, $n^{0.001}$ is larger than $\log n$. But that does not make sense. $0.001$ is close to $0$. So anything raised to the power $0.001$ should come out to be slightly more than $1$ right?
e.g if n = 1000,000 then $n^{0.001}$ is equal to $1.014$ whereas $\log n$ will be equal to almost 20 if $\log$ has a base of 2.
Where am I thinking wrong?
Is there any other way of showing this relationship?

Posting this as an answer mainly for space and typesetting. To piggyback off of dani_s' answer, consider the following table of values:
$\begin{array}{l | l}
n & \dfrac{\log n}{n^{0.001}} \\[10pt]
\hline
10^3 & 6.8602\\
10^{30} & 64.469\\
10^{300} & 346.208\\
10^{3000} & 6.90776\\
10^{30000} & 6.90776 \times 10^{-26}\\
\end{array}$
Since $n^{0.001}$ grows so slowly, it seems like $\log n$ grows faster. But for any $c>0$, $n^c$ will eventually beat $\log^k n$ for any $k$.
