# 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.

• If you want your two statements to look dual to each other, you need (b) to begin "if $\liminf \left|\frac{a_{n+1}}{a_n}\right| > 1$..." Of course, that's strictly weaker than the given statement. – Micah Sep 3 '19 at 21:28

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 $$\limsup_{n \to \infty} |\frac{a_{n+1}}{a_n}| = 2$$.
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.
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 $$\sum a_n <\infty$$.