Can a bounded number sequence be strictly ascending? [closed]

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 TaxelMar 16 at 15:14

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• A series deals with summation. A sequence deals with individual elements. – Subhasis Biswas Mar 12 at 12:19
• 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. – Teepeemm Mar 12 at 15:12

Yes.

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:

This is also relates to Zeno's Paradoxes.

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

Can a bounded number sequence be strictly ascending?

Sure it can.

Hint

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

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