Convergence of oscillating sequence I was trying to do a problem to prepare for my Real Analysis test but i got stuck. The problem states the following:
Suppose that $x_{n} \geq 0$ for all $n \in \mathbb{N}$ and that $\lim \left((-1)^{n} x_{n}\right)$ exists (Let's suppose $\lim \left((-1)^{n} x_{n}\right) = \alpha$). Show that $\left(x_{n}\right)$ converges.
What have i tried so far? My most promissing attempt went like this:
Define $z_{n}:=(-1)^nx_{n}$ We already know that $z_n \xrightarrow{\
} \alpha$. If we can prove that both subsequences $z_{2n}$ and $z_{2n - 1}$ converge to $\alpha$ we'll be done. For that, consider the following subsequence: $$z_{2n} = (-1)^{2n}x_{2n} = x_{2n}$$ We can say that $\lim x_{2n} = \alpha$ because $z_{2n} \xrightarrow{\
} \alpha$ as every subsequence of $z_{n}$ should converge to the same limit $\alpha$.
On the other hand, consider the subsequence: $$z_{2n-1} = (-1)^{2n - 1}x_{2n-1} = -x_{2n - 1}$$ We know that $\lim z_{2n-1} = \lim -x_{2n - 1} = -\lim x_{2n-1} = \alpha$. Therefore, $\lim x_{2n-1} = -\alpha$.
However, i wanted $$\lim x_{2n-1} = \alpha$$ for $\alpha$ bigger than zero in order to finish the proof, because then both odd and even subsequences would be converging to the same limit. Can anyone help me finish this exercise? If this idea cannot be fixed, can anyone suggest a solution?
Thanks in advance!
 A: Suppose $\{(-1)^n x_n\}$ is convergent to $x>0$.
$\implies \forall\  \epsilon \ \exists\  N_0\  \forall n>N_0,\ |(-1)^n x_n-x|<\epsilon $
$\implies \forall n>N_0,\ x-\epsilon< (-1)^n x_n<x+\epsilon$.
Choosing $\epsilon$ in such a way that $x-\epsilon>0$ we get that
$\forall n>N_0,\ 0<x-\epsilon< (-1)^n x_n$.
But this is impossible as for odd $m>N_0$, $(-1)^m x_m<0$. $\therefore x\not>0$
Similarly we can prove that $x\not<0$.
Therefore $x = 0$.Therefore $(-1)^nx_n \to 0$
$\implies \forall\  \epsilon \ \exists\  N_1\  \forall m>N_1,\ |(-1)^m x_m|<\epsilon/2,\exists\  N_2\  \forall n>N_2\ |(-1)^n x_n|<\epsilon/2 $.
So for $\forall m,n>\max(N_1,N_2), |(-1)^m x_m|<\epsilon/2, |(-1)^n x_n|<\epsilon/2$
Thus $\forall\  \epsilon \ \exists\  N\  \forall m,n>N,\ |x_m-x_n|<|x_m|+|x_n|<\epsilon $.
Therefore $x_n$ is cauchy. And since Cauchy sequences in $\mathbb R$ are convergent, so $x_n$ is convergent.
Now if $x_n \leq 0, Let z_n=-x_n\geq 0$, by above argument $z_n$ converges which in turn implies that $x_n$ converges.
