Does a convergent subsequence guarantee that the original sequence is also convergent? I am trying to get the intuition on the title statement or find a counterexample.
I know that if a sequence converges, then every subsequence of it converges as well.
But I am not sure that if a subsequence converges, then that guarantees that the original sequence also converges. My intuition says it's not true but I m struggling to find a counterexample.
 A: It does not.  A counterexample can be easily constructed by taking two convergent sequences, say $a_n$ and $b_n$, which converge to different limits, ie. $\lim_{n\rightarrow\infty} a_n = a \neq b = \lim_{n\rightarrow\infty} b_n$, and then creating a new sequence $c_n$ such that
\begin{equation}
c_n = 
\begin{cases}
a_n &\text{ if $n$ is even}\\
b_n &\text{ if $n$ is odd}
\end{cases}
\end{equation}
Here, the subsequence of even terms converges to $a$, and the subsequence of odd terms converges to $b$. Since $a\neq b$, $c_n$ does not converge. The proof follows from considering the definition of the limit of a sequence (the proof is not particularly difficult, and is a good exercise).
More complex sequences consisting of more than two convergent subsequences can also be created by this method (try it!).
A very simple example is provided by Bungo in their comment, which is the sequence
\begin{equation}
x_n = 
\begin{cases}
1 &\text{ if $n$ is even}\\
0 &\text{ if $n$ is odd}
\end{cases}
\end{equation}
The subsequences consisting of all $1$s and all $0$s converge to $1$ and $0$ respectively, whereas $x_n$ does not converge.

Also of interest to you might be the Bolzano-Weierstrass Theorem, which establishes that every bounded (and not necessarily convergent) sequence in $\mathbb R^n$ has a convergent subsequence. The proof of the statement is easy to follow and will likely give you a better idea of convergence of subsequences.
