might be a Taylor expansion question about $\log$ functions

Question

I was reading a paper which involves some results as: $$\begin{aligned} -\frac{n}{2}\log\left(1+\frac{1}{n}\right)+\frac{1}{2} & = \frac{1}{2n}\frac{n-\frac{2}{3}}{2n}=\frac{1}{4n}+o(n^{-2})\\ \frac{n-1}{n+1}+1-n\log\left(1+\frac{1}{n}\right) & =\frac{n-\frac{1}{2}+\frac{1}{6n}}{n+1}+o(n^{-3})\\ -\frac{n}{2}\log\left(1-\frac{1}{n^2}\right) & =\frac{1}{2n}+o(n^{-3})\\ -\frac{n}{2}\log\left(1-\frac{1}{n^2}\right)+\frac{n}{n+1} & =\frac{n+\frac{1}{2}+\frac{1}{2n}}{n+1}+o(n^{-3}) \end{aligned}$$

I tried the Taylor expansion about the $\log$ function around 1 for the first two, but I did not get the results as shown above.

Any ideas about this? Thanks!

Answer 1

hint

Near $x=0$, we have

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$

Near infinity,

$$\ln(1+\frac 1n)=\frac 1n-\frac{1}{2n^2}+\frac{1}{3n^3}+o(\frac{1}{n^3})$$

with $$o(\frac{1}{n^3})=\frac{1}{n^3}\epsilon(n)$$

and $$\lim_{n\to +\infty}\epsilon(n)=0$$

Answer 2

You can expand $$ \log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}+o(n^{-3}) $$ Therefore \begin{align} -\frac{n}{2}\log\left(1+\frac{1}{n}\right)+\frac{1}{2} &=-\frac{1}{2}+\frac{1}{4n}-\frac{1}{6n^2}+o(n^{-2})+\frac{1}{2}\\[6px] &=\frac{1}{2n}\frac{3n-2}{6n}+o(n^{-2})\\[6px] &=\frac{1}{2n}\frac{n-\frac{2}{3}}{2n}+o(n^{-2})\\[6px] &=\frac{1}{2n}\left(\frac{1}{2}-\frac{1}{3n}\right)+o(n^{-2})\\[6px] &=\frac{1}{4n}-\frac{1}{6n^2}+o(n^{-2}) \end{align} This is quite a contorted way, because the result is already in the first line. The term $-\frac{1}{6n^2}$ is missing.

Let's try \begin{align} -\frac{n}{2}\log\left(1-\frac{1}{n^2}\right) &=-\frac{n}{2}\left(-\frac{1}{n^2}-\frac{1}{2n^4}+o(n^{-4})\right)\\[6px] &=\frac{1}{2n}+\frac{1}{4n^3}+o(n^{-3}) \end{align} Again a term is missing.

Unless it's $O(n^{-3})$ instead of $o(n^{-3})$ and similarly for the first one.