Logarithmic asymptotic formula using o(1)

Question

Let's say that we have $\lim\limits_{n\rightarrow\infty} \frac{1}{n}\log a_n \, = \, a$. My professor said that this is equivalent to $a_n \sim \exp(na)$ for $n$ large. Intuitively it makes sense but I cannot really do the algebra to get the result. I must show that $a_n = \exp(na)(1+o(1))$ for $n$ large, right ? Could someone help me with this ?

What I can do so far is to say that the limit above is equivalent to $\frac{1}{n}\log a_n \, - \, a = o(1)$ for $n$ large, which then gives $a_n= \exp(n(a+o(1))$ for $n$ large. How is my result related to the one my prof said ? is it weaker or is it the same ?

Answer 1

Considering that $a_n \sim \exp(na) \iff \lim_{n\to \infty} \frac{a_n }{\exp(na)}=1 $, the assertion is wrong:

$$ \frac{a_n}{\exp(na)}=\exp\left(n \left(\frac{\log a_n}{n}-a\right)\right)=e^{ng(n)}$$ with $g(n)=\frac{\log a_n}{n}-a$. You are told that $\frac{1}{n}\log a_n \, \to \, a \implies g(n)\to 0$, but you need more than that, you need $g(n)=o(1/n)$.