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Let $\{a_n\}$ be a sequence such that for all $n$, $a_n>0$. I have to prove that if $\lim_{n\to\infty} a_n a_{n+1} =0$ then $\lim_{n\to\infty} a_n=0$. Here's my reasoning: $\forall\varepsilon>0,\exists k$ such that $\forall n>k$ we have $|a_n a_{n+1}|=a_n a_{n+1} < \varepsilon$ and thus $a_n=|a_n|=|a_n-0| < \frac{\varepsilon}{a_{n+1}} \Rightarrow \lim_{n\to\infty} a_n=0$. However I don't feel comfortable with the $a_{n+1}$ on the r.h.s. What do you think?

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    $\begingroup$ This isn't true without some further assumptions. For example, take $a_n = 1$ for $n$ even and $a_{2k+1} = \frac{1}{k}$. $\endgroup$ – rogerl Mar 28 '15 at 18:56
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I don't think it holds. For example, Let $a_{2n}=\frac 1n,a_{2n+1}=\sqrt n$.

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  • $\begingroup$ Thank you. Indeed there must be a mistake. By the way, what was the flow in my reasoning? $\endgroup$ – user227087 Mar 28 '15 at 19:14
  • $\begingroup$ @user227087 In fact, I don't understand your reasoning...Why $a_n< \frac{\varepsilon}{a_{n+1}}$ implies $\lim_{n\to\infty} a_n=0$? $\endgroup$ – Censi LI Mar 28 '15 at 19:20
  • $\begingroup$ Well, because we can control the $\varepsilon/a_{n+1}$ by picking any $\varepsilon$ that we like. So the term $\varepsilon/a_{n+1}$ itself is like a "$\varepsilon$" in any regular proof of limit. $\endgroup$ – user227087 Mar 28 '15 at 19:24
  • $\begingroup$ @user227087 Well, I think to prove the existence of limit, you should fix $\varepsilon$ each time, but not vary it to control $a_n$. $\endgroup$ – Censi LI Mar 28 '15 at 19:38
  • $\begingroup$ @CensLI - but we need $a_n<\varepsilon'$ for all $\varepsilon'$. What's important is the $\varepsilon'$ which is indeed fixed. Once it is fixed, we can find $\varepsilon$ that will satisfy $\varepsilon/a_{n+1}<\varepsilon'$ and thus $a_n<\varepsilon'$. $\endgroup$ – user227087 Mar 28 '15 at 20:44
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Hint $a_{2n}=\frac{1}{2n}$ and $a_{2n+1} =\sqrt{2n+1}$.

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As said that's not correct. I think you should better argue by contradiction that $\lim_{n\to\infty}a_n\neq 0$. WLOG the limit exists and $$\lim_{n\to\infty} a_n=\lim_{n\to\infty}a_{n+1}=l>0\Rightarrow \lim a_na_{n+1}=l^2>0$$ in contradiction which means that necessarily $\lim_{n\to\infty}a_n=0$ .

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if limit exist then $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}a_{n+1}$ thus $\lim_{n\to\infty}a_na_{n+1}=\lim_{n\to\infty}a_{n}^2=0$ thus $\lim_{n\to\infty}a_n=0$ if limit does not exist we cannot comment on $\lim_{n\to\infty}a_n$

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As others have said, this property doesn't hold. The flaw in your reasoning is the inequality that ${\varepsilon \over a_{n+1}}<\varepsilon$, you're assuming $a_{n+1}>1$, which you don't necessarily know. You can have $a_k<1$ for infinitely many $k$, which is what goes wrong with the examples given by other posters.

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The definition of limit says: $\left ( \forall \varepsilon ' >0 \right )\left ( \exists N \right )\left ( \forall n>N \right )\left ( |a_n -L|<\varepsilon \right )$. Let $\varepsilon '>0$ be given. If you fix some natural number $k$, you can set $\varepsilon = \left \lfloor\varepsilon ' |a_k| \right \rfloor$ to obtain an $N$ such that for all $n>N$: $|a_na_{n+1}|<\varepsilon$ which gives $|a_n|<{\varepsilon \over |a_{n+1}|}$. But notice that we have the fixed subindex $k$ in the numerator and $n+1$ in the denominator, so we cannot conclude anything further.

What you cannot do is set $\varepsilon=\left \lfloor |a_n| \varepsilon' \right \rfloor$ for all $n$ since this isn't a number, it is a function of $n$. And if you were to try to work around this by taking, for example $\varepsilon =\displaystyle \inf _{n \in \mathbb{N}}\left \lfloor |a_n| \varepsilon' \right \rfloor$ you have no guarantee that this is a positive number.

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  • $\begingroup$ I don't understand why it matters if $a_{n+1}>1$ or not. Here's how my book proves the theorem $\lim_{n\to\infty}c a_n =c\lim_{n\to\infty}a_n$ for $c\neq 0$: $\forall\varepsilon>0,\exists N,\forall n>N:|a_n-L|<\frac{\varepsilon}{|c|}\implies|c||a_n-L|<\varepsilon\implies|ca_n-cL|=|c(a_n-L)|<\varepsilon$. We can go backwards in this case and here we have $\varepsilon/|c|$ even though it can be that $|c|<1$. How is that different from our case? Why $a_{n+1}$ must be greater than 1? $\endgroup$ – user227087 Mar 28 '15 at 19:36
  • $\begingroup$ You cannot conclude that ${\varepsilon \over a_{n+1}}<\varepsilon$ if you do not know that $a_{n+1}>1$. Does ${\varepsilon \over 0.5}<\varepsilon$? The theorem you brought up is different because $c$ is a fixed number, and as such we can take $|a_n-L|<{\varepsilon \over |c|}$ from the get go, and hence you get $|c(a_n-l)|<|c||{\varepsilon \over c}|<\varepsilon$. $\endgroup$ – Reveillark Mar 28 '15 at 19:42
  • $\begingroup$ By the way, I don't claim that $\varepsilon/a_{n+1}<\varepsilon$. I claim that $\varepsilon'=\varepsilon/a_{n+1}$ is an "epsilon by itself". Just like $\varepsilon/|c|$ in the previous example. $\endgroup$ – user227087 Mar 28 '15 at 19:42
  • $\begingroup$ You cannot take $|a_n-L|<{\varepsilon \over a_{n}}$ for all $n$ since the expression on the right depends on $n$, and is not fixed like ${\varepsilon \over c}$. $\endgroup$ – Reveillark Mar 28 '15 at 19:43
  • $\begingroup$ sorry for the dumb questions, but I really want to understand this. The definition of limit tells us that given any $\varepsilon'>0$ we must provide a number $N$ such that $\forall n>N: |a_n-L|<\varepsilon'$. In our case $\forall\varepsilon'>0$ we can give $N=k$ and by setting $\varepsilon=\lfloor\varepsilon'a_{n+1}\rfloor$ we have $|a_n|<(\lfloor\varepsilon'a_{n+1}\rfloor)/a_{n+1}<\varepsilon'$. Why is it wrong? How's the fact that $a_{n+1}$ isn't fixed changes the rules of the game? $\endgroup$ – user227087 Mar 28 '15 at 20:01

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