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 ?
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 $1$.
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.
- $\ln(xy) = \ln x+ \ln y$.
- 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 $s \in (0,1)$.