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Self learning Analysis and found the following exercise in Bartle's Elements of Real Analysis:

Let $X = (x_n)$ be a sequence of strictly positive real numbers such that $\lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1.$Show that for some $r$ with $0\lt r \lt 1$ and some $C \gt 0 $ we have $ \ 0 \lt x_n \lt Cr^n$ for all sufficiently large $n \in \Bbb N$. Use this to show that $\lim(x_n) = 0.$

Now I haven't proven any significant results concerning sequences and hence am required to solve this problem with basically nothing but the definition of a limit.

I am not getting too far with this because I cannot see a way to get there with a direct implication and I cannot for the life of me negate "for all sufficiently large $n$" in the context of the sentence above.

Would be extremely grateful if someone could help me out.

What I've done so far:

We know that $l = \lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1$. If $l \lt 0$ we can find a neighbourhood of $l$ entirely contained in $(-\infty, 0)$ which would contain no elements of the sequence $ \left({\frac {x_{n + 1}} {x_n}} \right) $ all of whose elements are positive leading to a contradiction. This implies $l \ge 0.$

We can pick $ 0 \lt \epsilon \lt 1 - l$. There is $m \in \Bbb N$ such that

$$n \ge m \implies \left|{\frac {x_{n + 1}} {x_n} - l}\right| \lt \epsilon \implies \frac {x_{n + 1}} {x_n} \lt \epsilon + l \lt 1 \implies x_{n + 1} \lt x_n$$

Also can someone tell me if $\lim (x_n) = 0$ necessarily? Doesn't $\left({2 + \frac 1 n}\right) $ satisfy the given conditions??

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    $\begingroup$ Re: your last question, for the sequence $x_n=2+1/n$ the ratio limit is exactly one. For the criterion to apply, you need it to be strictly smaller than that. $\endgroup$
    – E.P.
    Mar 3, 2014 at 11:41
  • $\begingroup$ @episanty: Alright. Thanks for the clarification. So I need to derive more than the fact that this is decreasing sequence. $\endgroup$
    – Ishfaaq
    Mar 3, 2014 at 11:43
  • $\begingroup$ @Ishfaaq: An easy way: $\limsup x_n^{1/n}\le \limsup x_{n+1}/x_n$ for sequence of positive numbers so if $\limsup x_{n+1}/x_n <1$. Then the series $\sum x_n$ converges by root test. So $(x_n) \to 0$ $\endgroup$ Mar 5, 2014 at 4:19
  • $\begingroup$ @Jose Antonio: Haven't gotten to defining the Limit Superior yet. Will check this out when I do.. $\endgroup$
    – Ishfaaq
    Mar 5, 2014 at 16:07

1 Answer 1

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You have successfully unpacked the definition of the limit into the form that will become useful. You have already shown that there exists $m\in\mathbb N$ such that for all $n\geq m$ you have $$\frac{x_{n+1}}{x_n}<l+\epsilon=:r<1. \tag1$$

All you need now is to put some recursion to work. This is the core of the criterion: if the limit of ratios of successive elements is smaller than one, this means that each element of the sequence is smaller than the previous one by some fixed percentage. If you go multiple steps out, this percentage will be compounded, which is what gets you the bound.

To make that precise, note that multiple applications of the bound (1) will give you

  • $x_{m+1}<x_m\,r,$
  • $x_{m+2}<x_{m+1}\,r<x_m\,r^2,$
  • $x_{m+3}<x_{m+2}\,r<x_m\,r^3,$

and so on, so that a trivial induction step will give you $$x_{m+k}<x_m\,r^k.$$ This is essentially the bound you were after, if you set $C=x_m/r^m$.

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  • $\begingroup$ Thanks loads. Problem solved! $\endgroup$
    – Ishfaaq
    Mar 3, 2014 at 11:51

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