Let $A=\{a_n\}$ is a strictly-increasing sequence of positive integer. The natural density of this sequence is defined by $\delta(A)=\lim_{n\rightarrow \infty} \frac{A(n)}{n}$ whenever the limit exists and where $A(n)$ is the number of elements of $A$ not exceeding $n$. Is there a strictly-increasing sequence of positive integer $A=\{a_n\}$ such that $\delta(A)$ does not exists?
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Yes. Start with $1$. Then omit enough integers to reduce the ratio below $\frac{1}2$. Then include enough consecutive integers to increase the ratio above $1-\frac{1}3$. Then omit enough to reduce it below $\frac{1}4$. Then include enough to increase it above $1-\frac{1}5$. Keep going.
answered Jul 4 '16 at 12:48
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4$\begingroup$ Even simpler: reduce the ratio below $\frac13$ and then increase it above $\frac23$. Repeat. $\endgroup$ – TonyK Jul 4 '16 at 12:53
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2$\begingroup$ @TonyK: Sure, but I wanted an extreme example. $\endgroup$ – Brian M. Scott Jul 4 '16 at 13:05
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Consider the sequence of integers with an odd number of decimal digits.
answered Jul 4 '16 at 12:47
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2$\begingroup$ For this sequence the $A(n)/n$ ratio would bounce between approximately $1/10$ and $9/10$, right? $\endgroup$ – CiaPan Jul 4 '16 at 14:22
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1$\begingroup$ A similar example is the sequence of integers whose first digit (in the usual decimal representation) is a $1$. $\endgroup$ – Jeppe Stig Nielsen Jul 4 '16 at 21:42
