Approximating series with a convergent series

Question

I am quite new to the use of little $o$ notation. Suppose that I have a sequence $a_n$ such that $a_n= e^{n(-\alpha+o(1))}$ where $\alpha>0$. It is clear that $$\sum_{n=0}^\infty e^{-n \alpha}$$ converges by a geometric series. What can one say now about $$\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty e^{n(-\alpha+o(1))}=\sum_{n=0}^\infty e^{-n\alpha+o(n)}.$$ I thought this may have diverged due to the $o(n)$ in the exponent but I am not too sure how this would work.

Answer 1

$a_n= e^{n(-\alpha+o(1))}$ means that $a_n= e^{n(-\alpha+b_n)}$ with a sequence $(b_n)$ satisfying $\lim_{n \to \infty} b_n = 0$. It follows that there is an index $N$ such that $b_n < \alpha/2$ for $n \ge N$.

Then $0 < a_n< e^{-n\alpha/2}$ for all $n \ge N$, which implies that $\sum_{n=0}^\infty a_n$ is convergent.