Theorem 8.25 (Ratio Test) - Proof of part b), Mathematical Analysis 2nd ed. - Apostol Theorem 8.25 (Ratio Test) Given a series $\sum a_n$ of nonzero complex terms, let
$$r=\liminf_{n\to\infty}\left |\frac {a_{n+1}}{a_n}\right|\qquad R=\limsup_{n\to\infty}\left |\frac {a_{n+1}}{a_n}\right| $$
b) The series $\sum a_n$ diverges if $r>1$.
Proof. To prove b), Apostol says, we simply observe that r>1 implies $|a_{n+1}|>|a_n|$ for all $n\ge N$ for some $N$ and hence we cannot have $\lim_{n\to\infty}a_n=0$.
How are the following two sentences related?


*

*The series $\sum a_n$ diverges if $r>1$.

*"we cannot have $\lim_{n\to\infty}a_n=0$".


Should 2 imply 1?
Thank you.
 A: We know that 
$$\lim_{n\to\infty} a_n=0$$
is a necessary, though not sufficient, condition for the convergence of the series $\;\sum a_n\;$ , so what Apostol says is obvious.
What you ask is unrelated: we can have that $\;\sum a_n\;$ diverges and still $\;r=1\;$ and $\lim_{n\to\infty}a_n=0\;$. For example, with the harmonic series.
A: (2) does imply (1), by a simple boundedness argument.
Let $a_n$ be a sequence such that $lim_{n\rightarrow\infty}{a_n}\neq 0$.  Then there exist countably many terms such that $a_n\geq b$ or $a_n\leq b$ for some $b\in\mathbb{R}\setminus \{0\}$.  Without loss of generality assume we have countably many terms with $a_n\geq b>0$ and let these terms be labeled as $b_n$ (note this is a subsequence of $a_n$).  Then, regarding the sum, we have
\begin{equation}
\sum_{n=0}^{\infty}{a_n}\geq\sum_{k-0}^{\infty}{b_k}\geq\sum_{i=0}^{\infty}{b}
\end{equation}
Ans since $b$ is constant, the final sum is unbounded, so $\sum_{n=0}^{\infty}{a_n}$ is unbounded.
Hope this helps.
