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The problem I'm having is with the logs. I go:

$$\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}} \cdot \frac{n-2}{n-1} \Big)$$

$$=\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}}\Big) \cdot \lim_{n \to \infty} \Big(\frac{n-2}{n-1} \Big)$$

and here I know that $$\lim_{n \to \infty} \Big(\frac{n-2}{n-1} \Big) = \lim_{n \to \infty} \Bigg(\frac{1-\frac{2}{n}}{1-\frac{1}{n}} \Bigg) = \frac{\lim_{n \to \infty} ({1-\frac{2}{n}})}{\lim_{n \to \infty} (1-\frac{1}{n})} = 1$$

However, I don't know how to do the equivalent for $$\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}}\Big)$$

I know that the numerator and denominator functions converge as $n$ grows, but I don't know how to compute the limit algebraically and show that it's also $1$.

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    $\begingroup$ Can you use L'Hospital Rule? $\endgroup$ – sinbadh Mar 17 '16 at 3:25
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    $\begingroup$ Could you try to subtract to 1? In that case, the numerator is reaching 0 while the denominator reaching inf, resulting as 0. Therefore, the limit of the term you asked is 1. $\endgroup$ – Cuong Mar 17 '16 at 3:26
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We can write $\log{(n+1)}$ as $$\log{(n(1+\frac{1}{n}))}=\log{n}+\log\left(1+\frac{1}{n}\right).$$

Now $\log\left(1+\frac{1}{n}\right)$ is bounded, so is insignificant compared to $\log{n}.$ So the limit of $$\frac{\log(n+1)}{\log(n)}=\frac{\log{(n)}}{\log{(n)}}+\frac{\log{\left(1+\frac{1}{n}\right)}}{\log{(n)}}$$ tends to $1+0=1$.

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  • $\begingroup$ By bounded, do you mean that $\log{(1 + \frac{1}{n})}$ approaches $\log{(1)}$ as $n \to \infty$ and therefore $0$? Ah, just saw your edit. Thank you, this is very helpful. $\endgroup$ – jeremy radcliff Mar 17 '16 at 3:30
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    $\begingroup$ Well, kind of, but what I meant exactly by bounded is that $\log{(1+1/n)}$ has a supremum when $n \geq 1$ (the sup will be $\log{2}.$) So $\frac{\log{\left(1+\frac{1}{n}\right)}}{\log{(n)}}$ is less than or equal to $\frac{\log{2}}{\log{n}}$ when $n\geq 1,$ so it tends to 0. $\endgroup$ – dhk628 Mar 17 '16 at 3:34
  • $\begingroup$ Ah didn't see that you had edited your comment, but hopefully my comment helps! $\endgroup$ – dhk628 Mar 17 '16 at 3:35
  • $\begingroup$ Yes, thanks again. This is very cool, I didn't know about the term supremum but it makes total sense. Essentially you know that the denominator can't reach a value greater than a certain constant no matter what $n$ is, so you can treat it as that constant to take the limit. $\endgroup$ – jeremy radcliff Mar 17 '16 at 3:37
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    $\begingroup$ Yup, that's essentially what a supremum is. It's defined to be the smallest maximum, so it can never attain a value greater than the supremum. I should also add that this works because log is always positive for values greater than 1, so $0\leq \frac{\log{\left(1+\frac{1}{n}\right)}}{\log{(n)}}\leq \frac{\log{2}}{\log{n}},$ and the RHS tends to 0, so the expression in the middle must tend to 0 as well (I think this is called the sandwich/squeeze lemma). $\endgroup$ – dhk628 Mar 17 '16 at 3:42
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I say, $\lim\limits_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}}\Big) = 1 $ In wich case, I must show that

$\forall \epsilon > 0, \exists N>0$ such that $n>N\implies |\Big( \frac{\log{(n+1)}}{\log{(n)}}\Big) -1|<\epsilon$

$|\Big( \frac{\log{(n+1)}-\log{(n)}}{\log{(n)}}\Big)|<\epsilon$

$|\Big( \frac{\log{(1+1/n)}}{\log{(n)}}\Big)| < 1/N <\epsilon$

Let $N > 1/\epsilon$

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