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*}

One way could be $$\ln { \left( \frac { n+{ \left( -1 \right)  }^{ n }\sqrt { n } +a }{ n+{ \left( -1 \right)  }^{ n }\sqrt { n } +b }  \right)  } =\ln { \left( n+{ \left( -1 \right)  }^{ n }\sqrt { n } +a \right) -\ln { \left( n+{ \left( -1 \right)  }^{ n }\sqrt { n } +b \right)  }  } =\\ =\ln { \left( n\left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  \right)  } -\ln { \left( n\left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  \right)  } =\\ =\ln { \left( n \right) +\ln { \left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  }  } -\ln { \left( n \right) -\ln { \left( 1+\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  }  } =\\ =\frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { a }{ n }  \right)  }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } }  \right) -\left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n } -\frac { 1 }{ 2 } { \left( \frac { { \left( -1 \right)  }^{ n } }{ \sqrt { n }  } +\frac { b }{ n }  \right)  }^{ 2 }+O\left( \frac { 1 }{ { n }^{ 2 } }  \right)  \right) =\\ =\frac { a-b }{ n } +\frac { { \left( -1 \right)  }^{ n }\left( a-b \right)  }{ n\sqrt { n }  } +O\left( \frac { 1 }{ { n }^{ 2 } }  \right) \\ $$
