$\lim_{n \to \infty} \left(\frac{\ln(n+1)}{\ln(n)}\right)$ Stolz-Cesaro?

Question

How to apply Stolz-Cesaro in order to evaluate this sequence as n approaches to $\infty$?

I got to $$\lim_{k \to \infty} \frac{\ln\left(\frac{k+2}{k+1}\right)}{\ln\left(\frac{k+1}{k}\right)}$$

Answer 1

I don't think Stolz-Cesaro is needed here: $$\lim_{n \to \infty}\frac{\ln(n+1)}{\ln(n)} = \lim_{n\to\infty} \frac{\ln n + \ln(1+\frac{1}{n})}{\ln n} = \lim\left(1 + \color{red}{\frac{\ln(1+\frac{1}{n})}{\ln n}}\right) = 1$$ since the $\color{red}{\text{red}}$ expression goes to zero (numerator $\to 0$ and denominator $\to \infty$)

Answer 2

Although Stolz-Cesaro is unnecessary as VIVID says, if you were to use Stolz-Cesaro, you can say the following.

By Stolz-Cesaro, we have

$$\lim_{n\to\infty}\ln\left(\frac{\ln(n+1)}{\ln(n)}\right)=\lim_{n\to\infty}\frac{\ln(\ln(n+1))-\ln(\ln(n))}{(n+1)-n}=\lim_{n\to\infty}\frac{\ln(\ln(n))}n=0.$$

Now, taking exponents of both sides, we obtain:

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