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1.Series $\sum a_n$ diverges if (i) there exists $N\in \mathbb{N}$ such that $n≧N ⇒ |a_{n+1}/a_n|≧1$

2.Series $\sum a_n$ diverges if (ii)$\limsup |a_{n+1}/a_n|>1$

Here, both 1&2 are true. It is easy to see that (ii) implies (i), but is the converse true? Is 2 more general than 1?

Plus, it's usual that root test is harder to apply than ratio test, but i think root test is easier when a sequence $\{a_n\}$ has infinite 0 terms, since $|a_{n+1}/a_n|$ cannot be defined. Am i right or is there a "trick" to avoid this?

EDIT: I just noticed that this post is wrong. Statement 2 is indeed false.

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  • $\begingroup$ Have you tried a concrete example to see, whether for a sequence $(a_{n})_{n \in \mathbb{N}}$ (i) and (ii) are equivalent (say $a_{n}=1$ for all $n \in \mathbb{N}$)? And are you sure that (ii) implies (i)? $\endgroup$ – Nils Matthes Sep 18 '12 at 10:07
  • $\begingroup$ @Nils I made a mistake. (ii) implies 'There exists a subsequence satisfies (i)'. However, still not sure if (ii)⇒(i) is false. $\endgroup$ – Katlus Sep 18 '12 at 10:17
  • $\begingroup$ I think it is false: take $(1,1/2,1,1/2,...)$. $\lim \sup a_{n+1}/a_{n}$ ought to be $2$, while $a_{n+1}/a_{n}$ alternates between $1/2$ and $2$. So I guess you really need the modified definition as in your comment. $\endgroup$ – Nils Matthes Sep 18 '12 at 10:22
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Take $a_n=1$ to see that the converse is not true. i) means that the sequence $\{|a_n|\}$ is increasing, but it has to converge to $0$ in order to have convergence of the series $\sum_{n=1}^{+\infty}a_n$, so $a_n=0$ for $n$ large enough. But we had to assume that $a_n>0$ in order to i) make sense.

i) $\Rightarrow$ ii) is not true: take $a_n=2^{(-1)^n}$ for example.

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  • $\begingroup$ Thank you! Can you suggest me tricks to avoid the case in my post too if there exists? $\endgroup$ – Katlus Sep 18 '12 at 10:28

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