Assume that $a<0$ and that for $N$ large we have $a_N= \frac{\exp(Na)}{N} (1+o(1)) + O(\frac{1}{N})$.
Can I write that $a_N$ is equal to $\frac{\exp(Na)}{N} (1+o(1))$ ? since both terms are of order $O(\frac{1}{N})$. My question must be easy but I am new to this notation and still it's very confusing.
According to definition $$a_n=O(b_n)\iff \limsup_{n\to \infty} \left|{a_n\over b_n}\right|<\infty\\a_n=o(b_n)\iff \lim_{n\to \infty} {a_n\over b_n}=0$$according to Big O notation
Your conclusion is wrong since ${\exp(aN)\over N}$ is an exponential decreasing term that vanishes much faster that $O({1\over N})$ (i.e. $O({1\over N})$ becomes dominant for $N$ sufficiently large) because $1+o(1)$ can't compensate ${\exp(aN)\over N}$ (you can think of $1\over \sqrt n$ or $1\over \ln n$ as $o(1)$ and $1\over n^2\sin{1\over n}$ as $O({1\over n})$). Then it is suitable to say that $$a_n=O({1\over n})$$here is an example: