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How can one prove $e^n$ and $\ln(n)$, modulo 1, are dense in $[0,1]$, for $n=2,3,4...$?

By dense is meant, for any $0, there is an integer $m$ such that $0

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They are not uniformly distributed, which would mean e.g. $\lim_{n\to\infty} \frac{|\{k For example between $e^M$ and $e^{M+1}$ there are about $e^{M+1}-e^{M+\frac12}$ numbers $n$ with $\ln(n)\bmod 1>\frac 12$ and $e^{M+\frac 12}-e^{M}$ numbers $n$ with $\ln(n)\bmod 1<\frac 12$. These counts differ by a factor of $\sqrt e$ and that will be the relative proportion the larger the range of $n$ one checks becomes.

But they are dense in $[0,1]$ and that is the property you are looking for (as reflected by the edit of the question).

For the logarithm: Let $\epsilon>0$ be given. Find $N$ such that $\frac1N<\epsilon$. Then $0<\ln(n+1)-\ln n<\frac1n<\epsilon$ for all $n>N$ (because the derivative of $\ln$ is the reciprocal). Therefore the numbers $\ln n\bmod1$ with $N hit every subinterval of length $\epsilon$.

For the exponential this is a bit more difficult.

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    @math.n00b. For the log's case, only consider $\{log(2^n)\}$,then you will get the density property.2016-06-30