Seeking an intuitive proof that the divisor function $d$ has $d(n) = o(n^\epsilon)$ for all $\epsilon > 0$ It is clear that the number of divisors, d(n) grows alot more slowly than the number line, hence it is intuitive that >0, d(n)=o(n^)
But is there an intuitive way to formulate the proof? We can translate it into a simple-sounding problem:
Suppose we have a group of drummers playing different beats. 
The first drummer sets the beat//
The second drummer hits his drum every second beat of the first drummer//
The third hits every third beat//
etc
Now the question is simply: what is the rate at which the drummers' beats coincide increases, relative to the beat set by the first drummer? 
This is the same as the divisor function. On the 6th beat the first second and third and sixth drummers play together. On the 12th beat the first, second third fourth sixth and twelfth players hit their drums at the same time.
Obviously, the rate at which the number of drum beats coincide increases a lot more slowly than the number of beats the first drummer has played. But how much more slowly?
 A: It follows from $$\pi(2k)\log 2k\ge \sum_{p\le 2k}\log p= \sum_{p^m\le 2k}\log p+O(k^{1/2}) \\\ge \sum_{p^m\le 2k}\log p \lfloor 2k/p^m\rfloor = \log {2k\choose k} +O(k^{1/2})\ge O(k)$$ that $$n\le e^{rk} \implies \sum_{p| n} 1\le C\frac{\log n}{\log \log n} \implies d(n)\le 2^{C\frac{\log n}{\log \log n}}=n^{c/\log\log n}=O(n^\epsilon)$$
A: Here's a way to make your intuition more quantitative. Imagine the drummers have been playing for a long time, which you estimate is around $n$ beats, and you're about to start listening. How many drummers will hit the first beat you hear? Drummer 1 will definitely hit it. Drummer 2 hits half of all the beats, drummer 3 hits a third of them, and so on, up through drummer $n$, and the later drummers haven't begun playing. So the average number of hits per beat that we'll hear over the next few beats is approximately $\frac 1 1 + \frac 1 2 + ... + \frac 1 n$, which is approximately $\log n$ (see harmonic series) and grows more slowly than any positive power of $n$.
This idea doesn't yield a proof of the statement you mentioned, though, because the divisor function can exceed this "local average", and the argument above doesn't help us bound that behavior.
