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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.

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

3 Answers3

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Can a bounded number sequence be strictly ascending?

Sure it can.

Hint

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

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

Robert Israel
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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:

enter image description here

This is also relates to Zeno's Paradoxes.

ntg
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