Show $\ln\left(\frac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\frac{a-b}{n}+\mathcal{O}\left(\frac{1}{n^2} \right) $ 
I would like to prove the following:
  $$\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b}  \right)=\dfrac{a-b}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) $$

My attempt
i tried this way but i didn't get what i want :
\begin{align*}
\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)&=\ln\left(\dfrac{1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}}{1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}} \right) \\
&= \ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right)^{-1} \right) \\
&= \ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1-\dfrac{(-1)^{n}}{\sqrt{n}}-\dfrac{b}{n}+\mathcal{O}\left( \dfrac{1}{n}\right)\right) \right) \\
&=\ln\left( 1+\dfrac{a-b}{n}-\dfrac{(-1)^{n}}{\sqrt{n}}-\dfrac{1}{n}+\mathcal{O}\left( \dfrac{1}{n}\right) \right)
\end{align*}
 A: Your approach until 
$$
\ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right)^{-1} \right) \\$$
is correct. 
After that, we use $\ln (1+x) = x - \frac {x^2}2 + \frac{x^3}3+O(x^4)$ with $|x|< 1 $. 
We have
$$
\ln \left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right) =\frac{(-1)^n}{\sqrt n} + \frac an - \frac12\left(\frac1n+ \frac{2(-1)^na}{n\sqrt n}\right)+\frac13 \left(\frac{(-1)^n}{n\sqrt n} \right)+O\left(\frac1{n^2}\right)
$$
By the same way, 
$$
\ln \left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right) =\frac{(-1)^n}{\sqrt n} + \frac bn - \frac12\left(\frac1n+ \frac{2(-1)^nb}{n\sqrt n}\right)+\frac13 \left(\frac{(-1)^n}{n\sqrt n} \right)+O\left(\frac1{n^2}\right)
$$
Subtracting the latter from the former, we have
$$
\ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right)^{-1} \right) =\frac{a-b}n - \frac{ (-1)^n(a-b)}{n\sqrt n} + O\left(\frac1{n^2}\right).$$
Therefore, the stated formula does not hold when $a\neq b$. 
A: One way could be $$\ln { \left( \frac { n+{ \left( -1 \right)  }^{ n }\sqrt { n } +a }{ n+{ \left( -1 \right)  }^{ n }\sqrt { n } +b }  \right)  } =\ln { \left( n+{ \left( -1 \right)  }^{ n }\sqrt { n } +a \right) -\ln { \left( n+{ \left( -1 \right)  }^{ n }\sqrt { n } +b \right)  }  } =\\ =\ln { \left( n\left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  \right)  } -\ln { \left( n\left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  \right)  } =\\ =\ln { \left( n \right) +\ln { \left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  }  } -\ln { \left( n \right) -\ln { \left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  }  } =\\ =\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } }  \right) -\left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } }  \right)  \right) =\\ =\frac { a-b }{ n } +\frac { { \left( -1 \right)  }^{ n }\left( a-b \right)  }{ n\sqrt { n }  } +O\left( \frac { 1 }{ { n }^{ 2 } }  \right) \\ $$
