<s>[INST] Compute the limit of $1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)$ 
Compute the limit: $$\lim_{n \to \infty} \left(1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)\right) $$

Can someone help me to solve this limit? I forgot how to manupulate fractions in limit calculus.
 [/INST] If you want to go beyond the limit itself, write $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=1-n \log\left(1+ \frac{2}{2n - 1}\right)$$ Now, remembering that, for small values of $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ relace $x$ by $\frac{2}{2n - 1}$ which makes $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\log\left(1+ \frac{2}{2n - 1}\right)=\frac{2}{2n - 1}-\frac 12 \left(\frac{2}{2n - 1} \right)^2+\cdots$$ Now, use long division to get $$\log \left(\dfrac{2n + 1}{2n - 1}\right)=\frac{1}{n}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ So, for large values of $n$ $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=-\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
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