Why is $\limsup_{n\rightarrow \infty} (1/n) \log(a_n) < 0 \Rightarrow \sum_{n=0}^{\infty} a_n < \infty$ true?

Question

Let $(a_n)$ be a real values series.

Why is the following implication true? $$\limsup_{n\rightarrow \infty} (1/n) \log(a_n) < 0 \Rightarrow \sum_{n=0}^{\infty} a_n < \infty$$

Answer 1

Let's assume $a_n > 0$ to avoid the problem of $\log 0$, and the necessity of taking the absolute value. If

$$\limsup_{n\to\infty} \frac{1}{n}\log a_n = \lambda < 0,$$

then there is an $n_0$ such that, for all $n \geqslant n_0$, you have

$$\begin{align} \frac{1}{n}\log a_n &< \frac{\lambda}{2}\\ \Rightarrow (a_n)^{1/n} &< e^{\lambda/2} < 1\\ \Rightarrow a_n < e^{n\lambda/2}, \end{align}$$

and hence

$$\sum_{n=n_0}^N a_n < \sum_{n=n_0}^N e^{n\lambda/2} < \frac{e^{n_0\lambda/2}}{1-e^{\lambda/2}}$$

for all $N \geqslant n_0$. Thus the sequence of partial sums is bounded, hence $\sum\limits_{n=0}^\infty a_n < \infty$.