How to Solve the Equation Involving Landau's o Notation?

[Bestfrog]

Find $a,b,c \in \mathbb{R}$ such that

$$e^{\sqrt{n+2}-\sqrt{n}} -1 + cos(n^{-4})=a+\frac{b}{\sqrt{n}} + \frac{c}{n}=o(\frac{1}{n})$$I put $e^{\sqrt{n+2}-\sqrt{n}} \longrightarrow 1$

and $-1 + cos(n^{-4})=-\frac{1}{2 \sqrt{n}} + o(\frac{1}{n})$, so maybe $a=1$ and $b=-\frac{1}{2}$ but it's wrong!

(The solutions must be $a=1$, $b=\frac{1}{2}$ and $c=\frac{13}{24}$)

Any hint?

 

[Ray Vickson]

Find $a,b,c \in \mathbb{R}$ such that

$$e^{\sqrt{n+2}-\sqrt{n}} -1 + cos(n^{-4})=a+\frac{b}{\sqrt{n}} + \frac{c}{n}=o(\frac{1}{n})$$I put $e^{\sqrt{n+2}-\sqrt{n}} \longrightarrow 1$

and $-1 + cos(n^{-4})=-\frac{1}{2 \sqrt{n}} + o(\frac{1}{n})$, so maybe $a=1$ and $b=-\frac{1}{2}$ but it's wrong!

(The solutions must be $a=1$, $b=\frac{1}{2}$ and $c=\frac{13}{24}$)

Any hint?

Putting $e^{\sqrt{n+2}-\sqrt{n}} \to 1$ is not useful as it stands: you need to know how close it gets to 1 for large $n$. It is more revealing to put
$$\sqrt{n+2} = \sqrt{n}\sqrt{1 + \frac{2}{n} } = \sqrt{n} \left( 1 + \frac{1}{2}\frac{2}{n} + \frac{1}{2} \frac{-1}{2} \frac{1}{2!} \frac{2^2}{n^2} + \cdots \right),$$
so
$$\sqrt{n+2} = \sqrt{n} + \frac{1}{\sqrt{n}} - \frac{1}{2} \frac{1}{n^{3/2}} + \cdots . $$
Therefore,
$$e^{\sqrt{n+2}-\sqrt{n}} = e^{n^{-1/2} - (1/2) n^{-3/2} + \cdots}$$
and you can continue the expansion from there.

 

[Ray Vickson]

Find $a,b,c \in \mathbb{R}$ such that

$$e^{\sqrt{n+2}-\sqrt{n}} -1 + cos(n^{-4})=a+\frac{b}{\sqrt{n}} + \frac{c}{n}=o(\frac{1}{n})$$I put $e^{\sqrt{n+2}-\sqrt{n}} \longrightarrow 1$

and $-1 + cos(n^{-4})=-\frac{1}{2 \sqrt{n}} + o(\frac{1}{n})$, so maybe $a=1$ and $b=-\frac{1}{2}$ but it's wrong!

(The solutions must be $a=1$, $b=\frac{1}{2}$ and $c=\frac{13}{24}$)

The formula c = 13/24 is wrong.

Any hint?