6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 21:28
10
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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 $\sum a_n <\infty$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.