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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.

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  • $\begingroup$ 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. $\endgroup$
    – Micah
    Sep 3, 2019 at 21:28

3 Answers 3

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

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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.

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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 $\sum a_n <\infty$.

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