Convergent Sequence with Finite Number of Terms Deleted is Convergent

Theorem
Let $\left({X, d}\right)$ be a metric space.

Let $\left \langle {x_k} \right \rangle$ be a sequence in $X$.

If $\left \langle {x_k} \right \rangle$ converges, the deletion of a finite number of terms will still result in a convergent sequence.

Similarly, if $\left \langle {x_k} \right \rangle$ diverges, the deletion of a finite number of terms will still result in a divergent sequence.

Proof
Suppose, the sequence $\left \langle {x_k} \right \rangle$ converges to $x \in X$, i.e. for every $\varepsilon > 0$ there is some index $N$ such that $d(x_n, x) < \varepsilon$ for $n \geq N$. The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.

For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence. This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).

Lemma 1
Let $\left( X,d \right)$ be a metric space.

Let $\langle x_k \rangle$ be a sequence in $X$.

If $\langle x_{n_k} \rangle$ is a subsequence of $\langle x_k \rangle$, then $k \leq n_k \forall k \in \N$.

Proof of Lemma
Note that $1\leq n_1 < n_2 < \ldots < n_k < \ldots$ by definition of a subsequence.

We will proceed with induction on $k$ to prove our claim.

(k=1)

We need to show $1 \leq n_1$, but this holds from the definition of a subsequence.

(k=k+1)

Suppose $k \leq n_k$.

Then $k+1 \leq n_k +1$

Also, by definition of a subsequence, $n_k < n_{k+1}$.

Therefore $n_k +1 \leq n_{k+1}$.

From transitivity, we can conclude $k+1 \leq n_{k+1}$.

Lemma 2
Let $\left( X,d \right)$ be a metric space.

Let $\langle x_k \rangle$ be a convergent sequence in $X$,

then every subsequence of $\langle x_k \rangle$ is convergent.

Proof of Lemma
Let $\langle x_k \rangle$ be a convergent sequence in $X$.

Then $\exists L \in X$ such that $x_k \to L$ as $k \to \infty$.

Let $\langle x_{n_k} \rangle$ be a subsequence of $\langle x_k \rangle$.

We will show that $x_{n_k} \to L$ as $n_k \to \infty$.

Let $\epsilon >0$. Then $\exists N \in \N$ such that

$d(x_k,L)< \epsilon$ for all $k \geq N$.

From Lemma 1, we have $n_k \geq k$, therefore

$d(x_{n_k},L) < \epsilon$ for all $n_k \geq k \geq N$.

Hence $x_{n_k} \to L$ as $n_k \to \infty$.

Also see

 * Tail of Convergent Series