0
$\begingroup$

Prove that if $\liminf \left|\frac{a_{n+1}}{a_n}\right|>1$, then the series $\sum a_n$ diverges.

What I did was: let $c\geq 1$

If $\liminf \left|\frac{a_{n+1}}{a_n}\right|>c \implies \exists n_0 \in \mathbb{N}$ such that |$a_{n+1}|>|a_{n}|c\quad \forall n>n_0$.

then $|a_{n+2}|>c \cdot |a_{n+1}|>c^2\cdot |a_{n}|$ and so $|a_{n+p}|>c^p\cdot |a_n|$.

That makes $$\sum_{n=0}^{\infty} |a_n|>\sum_{n=0}^{n_0} |a_n|+|a_{n_0}|\sum_{n=n_0}^{\infty}c^n$$ but $\sum c^n$ diverges since $c>0$, therefore $\sum |a_n|$ diverges.

The problem is, the theorem means $\sum a_n$ diverges, and this proof only shows that it doesn't converge absolutely. It still could be conditionaly convergent. What can I do?

$\endgroup$
1
$\begingroup$

This is false. The series

$$\frac{1}{1^2} + \frac{2}{1^2} + \frac{1}{2^2} + \frac{2}{2^2} + \frac{1}{3^2} + \frac{2}{3^2} + \cdots $$

converges, and $\limsup \frac{a_{n+1}}{a_n} = 2.$ Perhaps you meant $\liminf \left |\frac{a_{n+1}}{a_n}\right | >1.$

$\endgroup$
  • $\begingroup$ Very nice example. Thanks for the correction. $\endgroup$ – Jacky Chong Oct 9 '16 at 2:54
  • $\begingroup$ God you're right! It was liminf! Sorry! (still, my proof for absolute convergence is correct for liminf and so the idea of $\lim a_n \neq 0$). I'll edit the question. Thank you. $\endgroup$ – Matheus barros castro Oct 9 '16 at 3:15
  • $\begingroup$ Ok I'm tired. @Matheusbarroscastro $\lim \inf_n |a_{n+1}/a_n| > 1$ means that $|a_n|\to \infty$ $\endgroup$ – reuns Oct 9 '16 at 3:48
  • $\begingroup$ Small question. Who down voted this and why? $\endgroup$ – zhw. Oct 9 '16 at 5:05

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.