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Is the sequence $(x_n)_{n=1}^\infty$, where $x_n$ is the fractional part of $1+\frac{1}{2}+\dots+\frac1n$, dense in $(0,1)$? The fractional part of a number $y$ is defined as $y-\lfloor y\rfloor$.

For a sequence like $a,2a,3a,\dots$ where $a$ is an irrational number, it is known that the fractional part sequence is dense. (I think there's even a name for this result, but I can't recall.) The proof uses a pigeonhole-style argument to show that the sequence must fall into any small interval of $(0,1)$ and relies on the linearity of the sequence, which we don't have in our sequence.

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  • $\begingroup$ The fractional part of $a,2a,\dots$ is dense is due to the equidistribution theorem. $\endgroup$ – Mark May 8 '17 at 6:40
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The sequence is divergent. Let $N>0$. After the $N$-th term, the succesive terms are $<1/N$ apart. Eventually the sequence passes an integer $K$ and later $K+1$. Every point in the interval $[K,K+1]$ is then within $1/N$ from an $x_n$. So the fractional parts of $x_n$ are dense in $[0,1]$.

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  • $\begingroup$ This actually proves a stronger claim: that if $a_n\to 0$ and $a_n>0$ and $\sum a_n$ diverges, then the fractional part of the sum is dense in $[0,1]$. $\endgroup$ – 5xum May 8 '17 at 7:10
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Yes it is dense

Consider splitting $[0,1)$ into sub-intervals $\left[0,\frac1m\right)$, $\left[\frac1m,\frac2m\right)$, $\ldots$, $\left[\frac{m-1}m,1\right)$ for arbitrary integer $m$

Now look at $x_n$ for $n=m,m+1,\ldots, 4m$. This will loop round at least once as $\frac1{m+1}+\frac1{m+2}+\cdots +\frac1{2m}+\frac1{2m+1}+\cdots +\frac1{4m} \gt m\times \frac1{2m}+2m\times \frac1{4m}=1$ (alternatively $\log(4m)-\log(m)=\log(4)>1$) and will hit each sub-interval at least once since each step is less than $\frac1m$. So no sub-interval will be missed.

Since $\frac1m$ can be made arbitrarily small, this results in $(x_n)_{n=1}^\infty$ being dense in $[0,1)$

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