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) $

Question

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*}

Answer 1

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 $a\neq b$.

Answer 2

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