$\lim_n ((\log(n+1)/(n+2))/(\log(n*(n+2)/(n+1)^2))- n$

Question

im kinda new in the field of mathematics, and i would really appreciate some help on the limit $$ \lim_{n\to\infty} \frac{\log\frac{n+1}{n+2}}{\log\frac{n(n+2)}{(n+1)^2}} - n. $$ Thank you all for the attention! Have a good one!

Answer 1

The expression under limit can be written as $$\frac{\log(1+a)}{\log(1+b)}-n$$ where $$a=-\frac{1}{n+2},b=-\frac{1}{(n+1)^2}, \frac{a^2}{b}\to - 1$$ and this can be further expressed as $$\dfrac{\log(1+a)-a+a-n\log(1+b)}{b\cdot\dfrac{\log(1+b)}{b}}$$ and clearly the denominator can be replaced by $b$. Next we split the fraction in two parts as $$\frac{\log(1+a)-a}{a^2}\cdot\frac{a^2}{b}+\frac{a-n\log(1+b)}{b}$$ The first part tends to $1/2$ and the second part tends to $0$ as shown next. Split second part as $$\frac{a-nb} {b} +\frac{b-\log(1+b)}{b^2}\cdot nb$$ and both of these tend to $0$. The key here is the limit $$\lim_{x\to 0}\frac{\log(1+x)-x}{x^2}=-\frac{1}{2}$$ which can be established either via Taylor series or via L'Hospital's Rule or using a complicated analysis starting from the definition $$e^x=\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n$$ and proving the limit $$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$