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I'm trying show that if $x \in (0,1)$, then sequence $\{x^n\}_\mathrm{n=1}^\infty$ converges with limit $0$.

I think the way to go about this is to show it has a limit, then show the limit equals $0$.

I honestly don't know what I should assume when starting this proof. We know $x \in (0,1)$ for certain. I could show that $\{x^n\}$ gives $x^\mathrm{n_1} > x^\mathrm{n_2}$ for $n_1 < n_2$ for an $x \in (0,1)$.

At a certain point, you want to say "Look! It can't go any lower than just above $0$ if the exponent goes to $\infty$. Clearly it's bounded!"

Basically, the proof and what to assume/use when looking at the initial information would be great.

Bob Mcdonald
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1 Answers1

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Hint :

1) Show that the sequence is decreasing and bounded below by $0$.

2) Deduce that the sequence converges to a limit $l$.

3) From the fact that $x^{n+1} = x \times x^n$, deduce that $l = x \times l$, and then that $l=0$.

TheSilverDoe
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