Power of little o - asymptotic - series Consider the following, when $n \to + \infty$:
$$\log\left(1 + \frac{-1}{q}n^{\frac{1-q}{q}} +o(n^{-1}) \right)= R_1(n^{\frac{-1}{q}}) + o(n^{-1}).$$
And I am trying to determine the polynomial $R_1$. From a lemma, I know that this polynomial satisfies $R_1(0)=0$ and is of degree at most $q$, where $q$ is an integer $\geq 3$.
My problem is with the little-o notations, I don't see how to work with them.
I used the Taylor expansion of $\log(1+x)$ and have the following expression for the left-hand side :
$$\sum_{j=1}^{\infty} (-1)^{j+1} \frac{1}{j} \left(\frac{-1}{q} n^{\frac{1-q}{q}} + o(n^{-1})\right)^j $$
Can anybody help ? Thanks a lot inadvance !

If you expand the logarithm using $\log(1+w)=w+\mathcal{O}(w^2)$, you find
\begin{align*}
& - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)\left( { - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right)} \right)^2 
\\ &
 =  - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)\left( {\frac{1}{{q^2 }}\frac{1}{{n^{2 - 2/q} }} + o\!\left( {\frac{1}{{n^{2 - 1/q} }}} \right) + o\!\left( {\frac{1}{{n^2 }}} \right)} \right)
\\ &
 =  - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)o\!\left( {\frac{1}{n}} \right) =  - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right).
\end{align*}
since, as $q\geq 3$,
$$
\frac{1}{{n^{2 - 2/q} }} \le \frac{1}{{n^{2 - 2/3} }} = \frac{1}{{n^{1 + 1/3} }} = o\!\left( {\frac{1}{n}} \right)
$$
and
$$
\frac{1}{{n^{2 - 1/q} }} \le \frac{1}{{n^{2 - 1/3} }} = \frac{1}{{n^{1 + 2/3} }} = o\!\left( {\frac{1}{n}} \right).
$$
