Prove $\limsup_{n \rightarrow \infty} a_n\geq1$ if $\lim_{n \rightarrow \infty} a_na_{n+1}=1$

Question

Let $\{ a_n \}_{n=1}^{\infty}$ be a bounded sequence s.t. $a_n>0,\ \forall n \in \mathbb{N}$. Suppose $\lim_{n \rightarrow \infty} a_na_{n+1}=1$, prove $\limsup_{n \rightarrow \infty} a_n\geq1$.

I tried to prove by contradicition using subsequence, but couldn't work it out.

Any help appreciated.

Answer 1

Hint: Suppose $\limsup a_n<1$, say $\limsup a_n=1-\delta$. Can you see why there is some $N$ such that $0<a_n<1-\delta/2$ for all $n\geq N$?

Answer 2

Since we work with positive sequences we have the inequality¹ $$1=\liminf a_n a_{n+1} \le \liminf a_n \limsup a_{n+1} \le \limsup a_na_{n+1} =1.$$ This means that $$\liminf a_n \limsup a_{n+1} =1.$$

So we get $$\liminf a_n = \frac1{\limsup a_{n+1}} = \frac1{\limsup a_{n}}.$$ So we see that $$\liminf a_n\cdot\limsup a_n=1.$$

Since we also knot that $\liminf a_n\le\limsup a_n$ we get that $$\liminf a_n \le 1 \le \limsup a_n.$$

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¹This is a special case of the more general fact that for positive sequences we have $$\liminf x_n \liminf y_n \le \liminf x_n y_n \le \liminf x_n \limsup y_n \le \limsup x_ny_n \le \limsup x_n \limsup y_n.$$

Probably you can find proofs at least of some (or maybe even all) of these inequalities on this site. But they can be definitely found in the book Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series. I have also mentioned this book in this answer to a question about one of these inequalities.

Answer 3

Hint: Suppose $|a_n a_{n+1}-1| \le \epsilon$, which, in particular, implies $a_n a_{n+1} \ge 1-\epsilon$. What does this tell you about at least one of the two terms $a_n$ or $a_{n+1}$?

Answer 4

If $\limsup a_{n}<1$ then $\forall n\geq n_{0}\; 0<a_{n}\leq \epsilon<1$ is true for some $\epsilon$ and some $n_{0}$.

This leads to $0<a_{n}a_{n+1}\leq \epsilon^{2}<1$ for $n\geq n_0$ contradicting that $\lim a_{n}a_{n+1}=1$.

Answer 5

You just need see if $\limsup (a_n) = c < 1$ then there are $n_0 \in \mathbb{N}$ such that $n > n_0 \rightarrow a_n \leq c$. So $a_na_{n-1} \leq c^2 < 1$, for $n > n_0$.

That is a contradiction to $\displaystyle \lim_{n \to \infty} a_na_{n+1} = 1$.

Sorry for my english...