How to show divergence of $\sum_{n = 1}^{\infty} 2^{-o(1)}\frac{1}{n}$?

Question

I am trying to show the divergence of the following sum:

$$\sum_{n = 1}^{\infty} 2^{-o(1)}\frac{1}{n}$$

where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:

since $o(1)$ means that $\lim_{n \rightarrow \infty} f(n)/1 \rightarrow 0$, then we can potentially look at $\max_n f(n)$ and $\min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!

Answer 1

If $f(n)=o(1)$, by definition, for some $\epsilon>0$, you will find an $N$ such that

$$n>N\implies|f(n)|<\epsilon\implies 2^{-f(n)}>2^{-\epsilon}.$$

Then the tail of the sum diverges as it is bounded below by an harmonic series.