Doubt regarding: Does existence of limsup of a sequence $x_n$ imply its convergence? Here is the proof that it does: Let's consider only bounded sequences. 
Let ($x_n$) be a bounded sequence then I am using the following definition for limsup of the sequence ($x_n$): 
limsup($x_n$)= inf $V$=inf{$v: x_n >v$ for at most a finite natural number $n$}=$x^*$, say. Consider the following:

$(a) \Rightarrow (b)$: Since $x^*$ is inf $V$ hence for any $\epsilon \gt 0 , \exists v\in V:$ $x^* \le v\lt x^*+\epsilon \implies x^*\in V \implies $ there can be atmost a finite number of $n\in N$ for which $x^*+\epsilon\lt x_n$. $\forall \epsilon \gt 0  $ $ x^*-\epsilon \notin V$. Hence $x_n \gt x^* -\epsilon$ for infinitely many $n$.  
Now my claim is that $(b)$ implies that $x_n$ is convergent, below is the proof: 
From $(b)$ above, let $ n=K $ (this K either exists or doesn't exist because of "atmost" condition) be the number for which $x^*+\epsilon\lt x_n$. 
(A) $\forall n\ge K$, we have $x_n \lt x^* +\epsilon$
 (B) Now we know that $x_n \gt x^* -\epsilon$ for infinitely many $n \implies \exists M \in N : \forall n \ge M$, we have $x_n \gt x^* -\epsilon$ 
From (A) and (B), for $n \ge L$=sup{$K,M$}, we have 
$x^* -\epsilon \lt x_n \le x^* +\epsilon \Rightarrow |x_n-x^*|\lt \epsilon \implies$ $lim (x_n)=x^*$
 
I know that if $X_n=$ sup{$x_k: k\ge n$}, then $lim (X_n)=x^*=$ inf{$X_m, m\in N$}. That's why the proof above doesn't seem right to me. I am struggling to know what is wrong in the above proof. Please help. Thanks in advance.

The implication you made in statement (B) is false. Just because there are infinitely many terms of the sequence that satisfy that property does not mean that they must all be consecutive.
