# Rate of convergence of an exponential function

If I have a function $$f = \exp(\sqrt{n} \cdot \frac{\sqrt{\log{n}}}{\sqrt{n}-\sqrt{\log n}}),$$ I can notice, that $$\lim_{n \to \infty} f = \infty,$$ but also I can notice that it goes very slowly to infinity, for example for $$n = 1000, f(n) \approx exp(2.87),$$ $$n = 10000, f(n) \approx exp(3.13).$$
How can I write that the rate of convergence is very slow? I mean I don't want to just write that $f \to \infty$, but I want use more accurate asymptotic notation which will help me to express that?

• Have you heard of the big-O notation? – user122283 May 31 '14 at 17:20
• it goes to infinity linearly. $$f(n)\sim n$$ – Jika May 31 '14 at 17:23
• And even, $f(n)\geqslant n$ for every $n$. – Did May 31 '14 at 17:26
• @QiaochuYuan Yes, you're right. I made two mistakes writing my functions here. – Rop May 31 '14 at 18:45
• @Did I made a mistake, while writting my formula here. Now it is what I meant. – Rop May 31 '14 at 18:45

I guess one way of seeing it rewriting the function $\frac{\sqrt{n} \log n }{\sqrt{n}-\log n} = \frac{\log n }{1 -\frac{\log n}{\sqrt{n}}}$. Hence your function becomes $n^\frac{1}{1-\frac{\log n }{\sqrt{n}}} \sim n^{1 +\frac{\log n}{\sqrt{n}} + O\big(\frac{1}{n^2}\big)}$, where each term $n^{a_n} \to_n 1$. Hence your expression grows $\sim n$ for large enough $n$.