If $(a_n)$ is any real sequence , then $(\frac{a_n}{1+|a_n|})$ has a convergent subsequence Consider the following two statements :
$S_1$: If $(a_n)$ is any real sequence , then $(\frac{a_n}{1+|a_n|})$ has a convergent subsequence .
$S_2$ : If every subseqeunce of $(a_n) $ has convergent subsequence , then $(a_n)$ is bounded . 
Which is of the follwoing statement is true ?
(A) Both $S_1$ and $S_2$ are true .
(B) Both $S_1$ and $S_2$ are false .
(c) $S_1 $ is false but $S_2$ is true .
(D) $S_2$ is false and $S_1$ is true .
My work : $1+|a_n|\geq 1+a_n>a_n$ . So $|(\frac{a_n}{1+|a_n|})|\leq 1$
. So it is a bounded sequence and via Bolzano-wietrass theorem it has a convergent subsequence . 
$a_{2n}$ and $a_{2n-1}$ are convergent . So they are bounded . Let the bounds be $M $ and $L$ respectively . 
Hence $a_n \leq M+L $ .So $a_n$ is bounded . 
So both $S_1$ and $S_2$ are true .   
 A: You are right about $S_1$, and your proof is correct.
Your proof for $S_2$ is wrong, since $(a_{2n})$ and $(a_{2n+1})$ does not represent every subsequence. 
Plus, they can not be bounded in general, since $(a_n)$ can be anything.
What you can do instead:
If $(a_n)$ is not bounded, then there exist a subsequence $(b_n)$ such that
$$\vert b_n\vert \to \infty.$$
And this sequence $(b_n)$ can not have a convergent subsequence.
So $S_2$ is true.
A: Your assessment of $S_1$ is correct: the statement is true.
Note that your approach for $S_2$ is incorrect: we cannot assume that the particular subsequences $(a_{2n}), (a_{2n-1})$ are convergent; all we know is that these subsequences have their own convergent subsequences in turn.
However, the statement is indeed true. I find it's easiest to prove by contrapositive: suppose that $(a_n)$ is not bounded.  Then for every $k \in \Bbb N$, there exists an $n_k \in \Bbb N$ such that $|a_{n_k}| > k$ (that is, we can construct a sequence with $|a_{n_k}| \to \infty$).  If we construct such a subsequence $(a_{n_k})$, then this subsequence can have no convergent subsequence (why is this the case?).
So, if $(a_{n})$ is such that every subsequence has a convergent subsequence, $(a_n)$ must be bounded.
A: Your work on the first statement looks good.
On the second, the problem statement says 
S2 : If every subseqeunce of $(a_n)$ has convergent subsequence , then $(a_n)$ is bounded.
So we don't know at $a_{2n}$ and $a_{2n+1}$ are convergent, only that they have a convergent subsequence.
