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?


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 $\liminf \left |\frac{a_{n+1}}{a_n}\right | >1.$

  • $\begingroup$ 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

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