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$(a_n)$ 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

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    $\begingroup$ 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. $\endgroup$
    – Martin R
    May 20, 2022 at 10:50

3 Answers 3

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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 $0$.

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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?

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$2$ ) 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 $\liminf a_n=0=\limsup a_n$

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