# $(a_n)$ is bounded and $a_n>0$ if $n$ is odd and $a_n<0$ if $n$ is even , does $a_n$ converge (including infinity) , and one more statement

$$(a_n)$$ is bounded and $$a_n>0$$ if $$n$$ is odd and $$a_n<0$$ if $$n$$ is even

1. does $$a_n$$ converge
2. $$\lim \limits_{n \to \infty}sup (a_n) \cdot \lim \limits_{n \to \infty}inf (a_n)<0$$

are the statements correct?

I believe the first one is not correct because I thought about the following sequence , let $$a_{n}=(-1)^n\cdot(-1)$$ so if $$n$$ is even we get $$(1) \cdot (-1) <0$$ and if $$n$$ is odd we get $$(-1) \cdot (-1) >0$$ similar to $$a_n=(-1)^n$$ the sequence diverges

for the second this is what I did:

$$\lim \limits_{n \to \infty}sup (a_n)=1=\lim \limits_{n \to \infty}(-1)^{2n-1}\cdot (-1)$$ and similar we have $$\lim \limits_{n \to \infty}inf (a_n)=-1=\lim \limits_{n \to \infty}(-1)^{2n}\cdot (-1)$$

so according to limits arithmetic properties $$\lim \limits_{n \to \infty}(-1)^{2n}\cdot (-1) \cdot \lim \limits_{n \to \infty}(-1)^{2n-1}\cdot (-1) = \lim \limits_{n \to \infty}(-1)^{4n-1}\cdot (1)<0$$

But this is wrong because I got that the second statement is correct and according to the book it is not correct

thanks for any help and tips and hopefully the translations are understandable

• You demonstrated only that (2) is correct for the sequence $a_n = -(-1)^n$. – For a counterexample to (2), try to find a sequence with the prescribed sign behavior that converges to zero. May 20, 2022 at 10:50

For the first question you're correct. As for the second one: Remember that $$\liminf(a_n)=\sup_{n\geq0}\inf_{m\geq n}a_m$$, So we can construct a sequence $$a_n$$ s.t $$a_{2n}<0$$ but $$\liminf(a_n)=0$$. An easy example is to take $$a_n=\frac{(-1)^{n+1}}{n}$$. It converges to $$0$$ so $$\liminf(a_n)=0$$ and therefore the product is $$0$$.

Your answer to the first one is right, the sequence $$a_n$$ you have is a counterexample to (1), showing it's false.

For (2), you only demonstrated that it's true for 1 particular sequence, not all of them.

Here's a hint: can you think of a sequence that fulfills the requirements of the exercise and also converges?

$$2$$ ) Consider $$(a_n)_{n\in \Bbb{N}}$$ where $$a_n=\frac{(-1) ^{n+1}}{n}$$

$$a_n>0$$ for $$n$$ odd and $$a_n<0$$ for $$n$$ even.

$$a_n\to 0$$ implies $$\liminf a_n=0=\limsup a_n$$