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How do I calculate the following limit with $\ln$?

$$\lim_{n\to \infty} \frac{\ln(n)}{\ln(n+1)}.$$

Would Taylor series expansion of $\ln(n)$ be a good place to start ?

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  • $\begingroup$ no you can't use taylor's expansion when the variable is more or less than 1 $\endgroup$ Mar 12, 2016 at 5:40
  • $\begingroup$ L'Hospital's rule $\endgroup$
    – zahbaz
    Mar 12, 2016 at 5:40
  • $\begingroup$ Taylor expansion of $\ln(1+x)$ for $x$ small is sort of helpful, seeing that $\ln(1+n)=\ln(n(1+1/n))=\ln n + \ln (1+1/n)$. $\endgroup$
    – ForgotALot
    Mar 12, 2016 at 5:41
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    $\begingroup$ you can even do it by replacing $n$ bye $1/x$ so when n will tend to infinity $1/x$ will trend to zero then taylor will be the beast!! $\endgroup$ Mar 12, 2016 at 6:00
  • $\begingroup$ One might also observe that $$ \ \frac{\ln n}{\ln (n+1)} \ = \ \log_{n+1} n \ \ , $$ so that this is equivalent to finding $ \ \lim_{n \to \infty} \ \log_{n+1} n \ $ . $\endgroup$ Mar 12, 2016 at 6:03

6 Answers 6

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We employ a direct approach relating $\log (n+1)$ to $\log n$. We have $$\log (n+1)=\log n+\log (1+1/n).$$

Therefore, for $n>1$, $$[\log n]/\log (n+1)=(1+[\log (1+1/n)]/\log n)^{-1}.$$ Since $\log n \to \infty$ and $\log (1+1/n)\to 0,$ we get the limit $1$.

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we know that L'hopital's rule is for limits in this form $\frac{\infty}{\infty}$ and you can see when n will tend to infinity $\ln (n)$ should also tend to infinity as we know this is a increasing function so use the hopital rule on differentiating numerator separately w.r.t $n$ we get $\frac{1}{n}$ and on differentiating denominator separately we get $\frac{1}{1+n}$

$$\implies\lim_{n\to \infty}\frac{1/n}{1/(1+n)}$$ $$\implies\lim_{n\to \infty}1+\frac{1}{n}$$ so therefore $1/n$ will tend to zero and the answer is 1

this question can also be solved by taylor's equation but for that you have to convert n to 1/x so when n will approach infinity 1/x will approach 0.

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  • $\begingroup$ L'Hospital is not applicable to sequences, only functions. $\endgroup$ Mar 12, 2016 at 12:37
  • $\begingroup$ @MathematicianByMistake how do you think this is a sequence? $\endgroup$ Mar 12, 2016 at 16:24
  • $\begingroup$ You are correct, my bad. If you wish, make some minor edit so I can upvote it again. $\endgroup$ Mar 12, 2016 at 16:35
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    $\begingroup$ now you can upvote !!!! $\endgroup$ Mar 12, 2016 at 16:41
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  1. $\ln(xy) = \ln x+ \ln y$.
  2. Write $n+1$ as a product $n+1 = n(1+1/n)$.
  3. Recall that $\ln n \to \infty$ as $n\to\infty$ and $\ln (1+1/n)\to 0$ as $n\to\infty$.

I hope this helps.

Also, you can write $\ln (n+1) = \ln n + (\ln (n+1) - \ln n)$.

The difference in parenthesis is $ \ln \frac{n+1}{n} = \ln (1+1/n)$, or can be estimated by the mean value theorem: it is equal to $1/(n+s)$ for some $s \in (0,1)$.

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Does l'Hopital work for you? $$\lim_{n\to\infty}\frac{\ln (n)}{\ln (n+1)}\overset{\text{L'H}}{=}\lim_{n\to\infty}\frac{1/n}{1/(n+1)}=\lim_{n\to\infty}\frac{n+1}{n}=\lim_{n\to\infty}1+\frac{1}{n} = 1.$$

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$$\lim\limits_{n\to\infty}\frac{\ln n}{\ln (n+1)}$$

$$\lim\limits_{n\to\infty}\frac{\ln n}{\ln \left(1+\frac 1 n\right)+\ln n}=\lim\limits_{n\to\infty}\frac{\ln n\big/\ln n}{\ln \left(1+\frac 1 n\right)\big/\ln n+\ln n\big/\ln n}=\lim\limits_{n\to\infty}\frac{1}{0+1}=\color{red}1$$

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  • $\begingroup$ Roper approach is to divide the numerator and denominator by $\ln n$ instead of just replacing $\ln(1+(1/n))$ with $0$. Such replacement of a sub-expression with its limit is invalid. $\endgroup$
    – Paramanand Singh
    Mar 12, 2016 at 17:17
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As already shown the more straightforward way is

$$\frac{\ln n}{\ln (n+1)} =\frac{\ln n}{\ln n+\ln \left(1+\frac1n\right)}\to 1$$

as an alternative by Cesaro-Stolz

$$\frac{\ln (n+1)-\ln n}{\ln (n+2)-\ln (n+1)}=\frac{\log\left(1+\frac1n\right)}{\log\left(1+\frac1{n+1}\right)}=\frac{n+1}{n}\frac{\log\left(1+\frac1n\right)^n}{\log\left(1+\frac1{n+1}\right)^{n+1}} \to 1\cdot \frac 1 1 =1$$

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