Prove that, if $\sum\limits_na_n$ converges, then $\sum\limits_n \frac{a_n}{a_{n+1}}$ diverges.
I believe it's true and even seems obvious, but once I tried to prove it I found it's not that trivial and I can't prove it rigorously.
Prove that, if $\sum\limits_na_n$ converges, then $\sum\limits_n \frac{a_n}{a_{n+1}}$ diverges
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sequences-and-series
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1Why do you think that it seems obvious? – 2017-01-17
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1@OpenBall I said that because I don't think $\frac{a_n}{a_{n+1}}$ and $a_n$ can converge to $0$ at the same time. But I can't formalize it. – 2017-01-17
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0Think of the ratio test. – 2017-01-17
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1@Jose27 But isn't the ratio test only for $a_n > 0$? There is no such condition here. – 2017-01-17
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1@Jose27 ratio test is not necessarily conclusive. – 2017-01-17
1 Answers
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If $\frac{a_n}{a_{n+1}} \to 0$, then there exists $N$ such that for $n\ge N$, $|a_n|<|a_{n+1}|$. As $|a_{N+1}|> |a_N|$, we get $|a_{N+1}|>0$, so for each $n\ge N+1$, $|a_n|\ge |a_{N+1}|>0$, contradicting $a_n \to 0$.
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0Thanks. Is there such a theorem that "If $\frac{a_n}{a_{n+1}} \to 0$, then there exists $N$ such that for $n\ge N$, $|a_n|<|a_{n+1}|$"? – 2017-01-17
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0That follows from the definition of limit (use $\epsilon=1$). – 2017-01-17
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0Sorry, I got it. Thanks! – 2017-01-17