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I am reading this note, which is to derive Gaussian distribution from Binomial. But I am confused by one step:

$$\ln p \\\sim ln\frac{2}{\sqrt{2\pi}}-\frac{1}{2}\ln n-\frac{1}{2}\ln(1-\frac{m^2}{n^2})-\frac{m}{2}\ln(1+\frac{m}{n})\\ \sim\ln\frac{2}{\sqrt{2\pi}}-\frac{1}{2}\ln n+\frac{m^2}{2n^2}+\frac{m^2}{2n}-\frac{m^2}{n}$$

That is, $\ln$$(1$$-\frac{m^2}{n^2})$ can be substituted by $\frac{m^2}{n^2}$. And it is explained as leading behavior. But I can't understand how it can be derived. Could you please help me with this problem?

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    Do you know the Taylor series expansion of the natural logarithm? – 2017-01-15
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    @B.Pasternak I know Taylor series expansion, but don't know Taylor series expansion of ln. I just googled it and know how it works. Thanks a lot! – 2017-01-15
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    Good. I'll post an answer so this doesn't stay unanswered. – 2017-01-15
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    @B.Pasternak Thanks for your work! – 2017-01-15

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The Taylor series expansion for the log is $log(1-x)=-x+\mathcal{O}(x^2)$ (as long as $|x|<1$), so $\frac{m^2}{n^2}$ is the leading behavior of $-\ln(1-\frac{m^2}{n^2})$.