# Logarithmic asymptotic formula using o(1)

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 ?

• I believe this is what your prof meant. By saying $a_n\sim \exp (na)$ he was saying that $a_n = \exp(na + no(1))$ and not that $a_n=\exp(na)(1+o(1))$. The latter is a stronger assertion and it's false. If you take $a_n= e^{na+\sqrt{n}}$ then $\frac{1}{n}\log a_n\rightarrow a$ but it's not satisfying that $a_n=\exp(na)(1+o(1))$. – Yanko Jan 13 at 11:30

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 $$g(n)=o(1/n)$$.