Krdytkyu

Prove: If $lim_{ntoinfty}s_n=s$ 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 $s=0$.

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.

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

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 $ge 0$.