The title says it. Can a bounded number sequence be strictly ascending / descending?

I have a problem that tells me the sequence of fractional parts $(\{nx\})_{n\geq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.


closed as off-topic by user21820, Song, Alex Provost, RRL, Parcly Taxel Mar 16 at 15:14

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  • 1
    $\begingroup$ A series deals with summation. A sequence deals with individual elements. $\endgroup$ – Subhasis Biswas Mar 12 at 12:19
  • 4
    $\begingroup$ You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $(\{nx\})_{n\ge1}$ is never strictly increasing. $\endgroup$ – Teepeemm Mar 12 at 15:12


Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"

So basically the sequence of the partial sums of e.g. a geometric series with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:

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This is also relates to Zeno's Paradoxes.

  • $\begingroup$ +1 for Zeno's paradox $\endgroup$ – Pere Mar 12 at 16:57

Can a bounded number sequence be strictly ascending?

Sure it can.


$0.9 \;,\; 0.99 \;,\; 0.999 \;,\; \ldots$


I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.


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