Is it true that if $\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1$, then $\sum a_n$ diverges? I am reading "A Course in Analysis vol.2" by Kazuo Matsuzaka.
There is the following theorem ("ratio test") in this book.

Let $a_n \neq 0$ for all $n$.
(a) If $\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum a_n$ converges absolutely.
(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n \geq N$ for some $N$, then $\sum a_n$ diverges.

Is the following statement false?
(b') If $\limsup\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1$, then $\sum a_n$ diverges.
 A: Unfortunately  it is false: consider the sequence $a_n$ defined as
$a_n=\frac{1}{n^2}, \mbox{ if n is odd}$ and $a_n=\frac{1}{n^3}, \mbox{ if n is even}$. The series $\sum_{k=0}^{\infty}$ is convergent but the subsequence of the ratios $r_k=\frac{a_{2k+1}}{a_{2k}} = \frac{(2k)^3}{(2k+1)^2}$ is divergent. 
A: No. Consider a sequence $(a_n)$ like $(\frac 1 {2^{3}},\frac 1 {3^{3}}, \frac 2 {3^{3}}, \frac 1 {4^{3}},\frac 1 {4^{3}}, \frac 2 {4^{3}},...)$ where the (n-1)-st block has $\frac 1 {n^{3}}$ repeated $n-1$ times followed by $\frac 2 {n^{3}}$. Then $\frac {a_{n+1}} {a_n}=2$ for infinitely many $n$ but $\sum a_n <\infty$.  
A: That is indeed wrong, a counter example is the sequence $(a_n)$
$$
 \frac 12, \frac 22, \frac 14, \frac 24, \frac 18, \frac 28, \ldots
$$
Here $\sum a_n$ is convergent, but $\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}| = 2$.
More generally you can take any convergent series $\sum c_n$ with $c_n \ne 0$ and then define $$
 a_{2n} = c_n, a_{2n+1} = 2c_n \, .
$$
Then $\sum a_n$ is convergent as well, but $\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}| = 2$.
