Suppose I want to calculate the Taylor-Young polynomial of a function $f(n)=\ln(1+\frac{n}{a^n})$ with $a>1$ to the order two. What bothers me is that I am unsure what should I put in the $o(?)$ part.
I would get something like, at point $0$:
$$f(n)= \frac{n}{a^n} - \frac{n^2}{2a^{2n}} + o(n^2)$$
But I am sure unsure if this is right. Is the $o(n^2)$ correct?
On
You state you are looking for a polynomial, but you are going in a direction where you have an expression that is not polynomial in $n$. To just compute the Taylor polynomial directly:
If $$f(x)=\ln(1+x/a^x)$$ then $$f'(x)=\frac{1}{1+x/a^x}\frac{1-x\ln(a)}{a^{x}}$$ and $$f''(x)=\frac{1}{1+x/a^x}\frac{-2\ln(a)+x\ln(a)^2}{a^{x}}-\frac{1}{\left(1+x/a^x\right)^2}\left(\frac{1-x\ln(a)}{a^{x}}\right)^2$$
So $f(0)=0$, $f'(0)=1$, and $f''(0)=-2\ln(a)-1$, and $$f(x)=x-\frac{2\ln(a)+1}{2}x^2+\operatorname{o}(x^3)$$
You have $\log(1+x)= x-\frac{x^2}{2} + o(x^2)$. Then, for $x=\frac{n}{a^n}$ you get: $$\log\bigg(1+\frac{n}{a^n}\bigg)=\frac{n}{a^n}-\frac{n^2}{2a^{2n}} + o\bigg(\frac{n^2}{a^{2n}}\bigg)$$
EDIT: as Alex M. pointed out, for $n \to 0$ we have $a^{2n} \to 1$. So $o(\frac{n^2}{a^{2n}}) = o(n^2)$