Proving that $\lim_{n\to\infty}s_n=0$

Question

Prove: If $\lim_{n\to\infty}s_n=s$ and $\{s_n\}$ has a subsequence $\{s_{n_k}\}$ such that $(-1)^ks_{n_k}\geq0$ then $s=0$.

Proof. Since $\{s_n\}$ converges to $s$ then so does its subsequence $\{s_{n_k}\}$ . Again, if it is convergent then $\liminf=\limsup$, but for $k$ even the subsequence assumes the form $s_{n_{2k}}$, instead for odd the form $-s_{n_{2k+1}}$. For these to be equal, for large $n$, both limits $s$ and $-s$ must be equal to $0$. Since $\liminf=\limsup=0$ then $s=0$.

Answer 1

Your proof is fine. Here a proof without $ \lim \inf$ and $\lim \sup$:

Suppose that $s \ne 0$. WLOG we assume thatb $s>0$. Since $s_{n_k} \to s$ as $ k \to \infty$, there is $K$ such that $s_{n_k}>0$ for all $k>K$. If $k>K$ and $k$ is odd, we get $(-1)^ks_{n_k}<0$, a contradiction.

Answer 2

$\lim_{n \rightarrow \infty} s_n=s.$

Then every subsequence converges to $s$.

Given:

$s_{n_k}$ converges to $s$, and $(-1)^k s_{n_k} \ge 0$.

1) $k =2m$ then $s_{n_{2m}} \ge 0$, and this subsequence of $s_{n_k}$ converges to $s$:

$\lim_{m \rightarrow \infty} s_{n_{2m}} =s \ge 0.$

Similarly;

2)$ k=(2m+1)$ then $s_{n_{2m+1}} \le 0$, and this subsequence of $s_{n_k}$ converges to $s$.

$\lim_{ m \rightarrow \infty}s_{n_{2m+1}}=s \le 0.$

Hence?

Used : If every term of a convergent sequence is $\ge 0,$ then the limit is $\ge 0$.