# Divergence of a sequence with floors and square roots $a_n = n\sqrt2 - [n\sqrt2] + n\sqrt3 -[n\sqrt3]$

Is there any specific approach to prove the divergence of a sequence? For example, I have this problem:

"Prove that the sequence $$(a_n)_{n \ge 1}$$, where $$a_n = n\sqrt2 - [n\sqrt2] + n\sqrt3 -[n\sqrt3]$$ is divergent"

(the [ ] stands for the floor function )

I tried to solve it by finding to subsequences that have different limits, but it does not seem to work this way.

• Do you know that $\{ n \sqrt{2} \pmod{1} \}$ is dense in $[0,1]$? Apr 1, 2019 at 13:59
• I did not know that. But still I cannot solve the problem (using your fact), neither prove what you said. Apr 1, 2019 at 14:26
• You should try to prove that, for any constant $c\in(0,1)$, there exist an infinite number of values of $n$ such that $n\sqrt2-\lfloor{n\sqrt2}\rfloor>c$ or equivalently $\{n\sqrt2 (\bmod 1)\}$ is dense in $[0, 1]$. Since you have a sum consisting of an infinite number of terms greater than some constant, the sum diverges. Apr 1, 2019 at 14:36
• I tried but I still cannot prove it this way. If you find a solution, will you please post it? Apr 1, 2019 at 14:56
• Hint: It is known that $\sqrt2$ is irrational, hence has non-repeating decimals. As $\{10^n\sqrt 2\}$ represents the tail of the number, starting at the $n^{th}$ decimal, it cannot tend to a constant.
– user65203
Mar 17, 2020 at 13:37

Since $$\sqrt{2}+ \sqrt{3}$$ is irrational, so is $$(a_n)$$. Consider the first decimal digit of $$(a_{10^m})$$. If $$(a_{10^m})$$ would converge, the first decimal digit must eventually become constant, say $$d\in\{0,\ldots,9\}$$, i.e. for all $$m\geq M$$ we have first decimal digit of $$(a_{10^m})$$ equals $$d$$. But the first decimal digit of $$(a_{10^m})$$ is the $$m$$-th decimal digit of $$\sqrt{2}+\sqrt{3}$$, so that would mean that the irrational number is $$d$$-periodic, contradiction. So $$(a_{10^m})$$ doesn't converge and so $$(a_{n})$$ doesn't converge either.
• There is a little flaw in this reasoning: $\{n\sqrt 2\}+\{n\sqrt 3\}\ne\{n\sqrt 2+n\sqrt 3\}$.