How do I calculate $\lim_{n\to \infty} \frac{\ln(n)}{\ln(n+1)}$? 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 ?
 A: 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 $1$.
A: 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. 
A: *

*$\ln(xy) = \ln x+ \ln y$. 

*Write $n+1$ as a product $n+1 = n(1+1/n)$. 

*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 $s \in (0,1)$. 
A: Does l'Hopital work for you?
$$\lim_{n\to\infty}\frac{\ln (n)}{\ln (n+1)}\overset{\text{L'H}}{=}\lim_{n\to\infty}\frac{1/n}{1/(n+1)}=\lim_{n\to\infty}\frac{n+1}{n}=\lim_{n\to\infty}1+\frac{1}{n} = 1.$$
A: 
$$\lim\limits_{n\to\infty}\frac{\ln n}{\ln (n+1)}$$

$$\lim\limits_{n\to\infty}\frac{\ln n}{\ln \left(1+\frac 1 n\right)+\ln n}=\lim\limits_{n\to\infty}\frac{\ln n\big/\ln n}{\ln \left(1+\frac 1 n\right)\big/\ln n+\ln n\big/\ln n}=\lim\limits_{n\to\infty}\frac{1}{0+1}=\color{red}1$$
A: 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
$$\frac{\ln (n+1)-\ln n}{\ln (n+2)-\ln (n+1)}=\frac{\log\left(1+\frac1n\right)}{\log\left(1+\frac1{n+1}\right)}=\frac{n+1}{n}\frac{\log\left(1+\frac1n\right)^n}{\log\left(1+\frac1{n+1}\right)^{n+1}} \to 1\cdot \frac 1 1 =1$$
