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I'd like to prove that $$\lim_{n\to\infty} \frac{\alpha + n}{\beta + n} = 1$$

I've worked out that $$\frac{|\alpha - \beta|}{|\beta + n|} < \epsilon$$ if $n < N$ for integers $n$ and $N$, but I'm not sure where to go from there.

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    $\begingroup$ What is this sequence $s_n$ you are referring to, and how is that related? (I.e. are $\alpha$ and $\beta$ constants?) $\endgroup$
    – mrf
    Nov 7, 2017 at 10:11

6 Answers 6

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write $$\frac{\frac{\alpha}{n}+1}{\frac{\beta}{n}+1}$$ and this tends to $1$ for $n$ tends to infinity

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  • $\begingroup$ Sorry, there was a typo in my question--I forgot to include the absolute value bars. Please let me know if that changes anything. $\endgroup$
    – user484604
    Nov 7, 2017 at 9:04
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    $\begingroup$ @Taliant It does not $\endgroup$
    – Rchn
    Nov 7, 2017 at 9:51
  • $\begingroup$ I was immediately worried about what impact different $\alpha$ and $\beta$ values would have. I'm glad your answer reminds me about one of the simple limit rules! $\endgroup$
    – Brian J
    Nov 7, 2017 at 16:13
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If $\alpha = \beta$ it is obvious. Otherwise, fix $\varepsilon > 0$ and look for $N \in \mathbb{N}$ such that $$\Big|\frac{\alpha+n}{\beta+n}-1\Big| < \varepsilon \text{ when } n > N. $$ You find immediately $$\Big|\frac{\alpha-\beta}{\beta+n} \Big| < \varepsilon \rightarrow \frac{|\alpha-\beta|}{\varepsilon} < |\beta+n| \leq |\beta|+n$$ So if you set $N > \frac{|\alpha-\beta|}{\varepsilon}-|\beta|$ you get the result.

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  • $\begingroup$ A fantastic and concise explanation, thanks. $\endgroup$
    – user484604
    Nov 7, 2017 at 9:20
  • $\begingroup$ Very nice solution. $\endgroup$ Nov 7, 2017 at 9:27
  • $\begingroup$ Glad that this helps. $\endgroup$
    – Gibbs
    Nov 7, 2017 at 9:49
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Note that $$ \frac{\alpha + n}{\beta + n} = \frac{\alpha - \beta + \beta + n}{\beta + n}\\ = \frac{\alpha - \beta}{\beta + n} + \frac{\beta + n}{\beta + n}\\ = \frac{\alpha - \beta}{\beta + n} + 1 $$ so by definition $\lim_{n\to \infty}\frac{\alpha + n}{\beta + n} = 1$ means that for any $\epsilon>0$, there is an $N\in \Bbb N$ such that as long as $n>N$, we have $$ \left|\frac{\alpha + n}{\beta + n} - 1\right|<\epsilon $$ However, by the above argument, this is inequality is equivalent to $$ \left|\frac{\alpha - \beta}{\beta + n}\right|<\epsilon $$ and you already know that there is an $N$ that works in this case. So we're done.

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  • $\begingroup$ Sorry, there was a typo in my question--I forgot to include the absolute value bars. Please let me know if that changes anything. $\endgroup$
    – user484604
    Nov 7, 2017 at 9:04
  • $\begingroup$ @Taliant Not at all (once you see that $\frac{|\alpha - \beta|}{|\beta + n|}$ and $\left|\frac{\alpha - \beta}{\beta + n}\right|$ are the same). In fact, that's what I've actually used in my answer, so if anything you've made it clearer how to put the two together. $\endgroup$
    – Arthur
    Nov 7, 2017 at 9:06
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You need to establish

$$|\alpha-\beta|<|\beta+n|\epsilon.$$

This is obviously obtained with

$$n>\frac{|\alpha-\beta|}\epsilon-\beta,$$ where the RHS is a well-defined finite number.

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Note that

$$ \left|\frac{\alpha+n}{\beta+n}-1\right| = \left|\frac{\alpha+n-\beta-n}{\beta+n}\right| = \left|\frac{\alpha-\beta}{\beta+n}\right| = \frac{|\alpha-\beta|}{|\beta+n|} $$

You showed that the right side is finally $<\epsilon$ for large enough $n$. This means that the left side will too. And this is the definition of

$$\lim_{n\to\infty} \frac{\alpha+n}{\beta+n}=1.$$

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If $\alpha=\beta$ it's trivial if not you can divide by $|\alpha-\beta|$ to get $$\frac1{|\beta+n|} <\frac\epsilon{|\alpha-\beta|} $$ Now take the greatest integer smaller or equal to $\beta+n$ denote it by $[\beta+n] $ then $\frac1{|\beta+n|} <\frac1{[|\beta+n|]} <\frac\epsilon{|\alpha-\beta|} $ the last inequality is due the Archimedian property.

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