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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.

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  • $\begingroup$ Do you know that $\{ n \sqrt{2} \pmod{1} \}$ is dense in $[0,1]$? $\endgroup$ Commented Apr 1, 2019 at 13:59
  • $\begingroup$ I did not know that. But still I cannot solve the problem (using your fact), neither prove what you said. $\endgroup$
    – Ddang
    Commented Apr 1, 2019 at 14:26
  • $\begingroup$ 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. $\endgroup$ Commented Apr 1, 2019 at 14:36
  • $\begingroup$ I tried but I still cannot prove it this way. If you find a solution, will you please post it? $\endgroup$
    – Ddang
    Commented Apr 1, 2019 at 14:56
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    $\begingroup$ 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. $\endgroup$
    – user65203
    Commented Mar 17, 2020 at 13:37

1 Answer 1

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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.

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  • $\begingroup$ There is a little flaw in this reasoning: $\{n\sqrt 2\}+\{n\sqrt 3\}\ne\{n\sqrt 2+n\sqrt 3\}$. $\endgroup$
    – user65203
    Commented Mar 17, 2020 at 13:40
  • $\begingroup$ @YvesDaoust I don't follow. The only difference is if the number is starting 1. or 0. and that's why I took the first decimal digit $\endgroup$
    – SK19
    Commented Mar 17, 2020 at 13:41

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