Proof that changing a finite number of terms in a series does not change where or not it converges I want to prove the following theorem:

Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge

I don't know how to begin as I am not sure how to translate that statement into symbols. Furthermore, I am not sure what "Changing a finite number of terms" means. Does it mean changing the value of a particular term or omitting it entirely or something else?
Please advise.
 A: Cauchy's criterion for convergence (assuming we're dealing with $\mathbb{R}^n$ or $\mathbb{C}^n$, technically a complete space in complete generality) says that a series $\{s_n\}_{n=0}^{\infty}$ with $s_n = a_0 + a_1 + \cdots + a_k$ converges if and only if for every $\epsilon > 0$, there is an $N > 0$ so that for all $m > n \geq N$, we have
$$
\left| s_m - s_n \right| = \left| \sum_{k = 0}^m a_k - \sum_{k=0}^n a_k \right| = \left|\sum_{k=n+1}^{m} a_k\right| < \epsilon.
$$
The key thing to realize is the "there is an $N > 0$ so that for all $m > n \geq N$." This is basically saying that it doesn't matter what happens for a finite number of terms before we get to a certain cutoff point.
So, suppose we change a finite number of terms in a series and say that $K$ is the largest $a_K$ that we change in the series. If we take $N_0 > \max(K,N)$, where $N$ is given as in the equivalence above (since we know the original series converges), we still obtain
$$
\left| \sum_{k = 0}^m a_k - \sum_{k=0}^n a_k \right| = \left|\sum_{k=n+1}^{m} a_k\right| < \epsilon.
$$
for all $m > n \geq N_0$. Therefore, the convergence of the series is not affected in the sense that one converges if and only if the other does also. However, they may in fact converge to different limits because the Cauchy Criterion does not give any information about the limit itself.
A: Suppose $x_n \to x$. This means for all $\epsilon>0$ there exists some $N$ such that if $n \ge N$, then $|x_n-x| < \epsilon$.
Now suppose $x'_n$ is a sequence such that for $n \ge M$, then $x'_n = x_n$.
Let $\epsilon>0$, there exists some $N$ that 'works' for the original sequence $x_n$. Now take $N'= \max(N,M)$. Then if $n \ge N'$, we have $|x'_n -x| < \epsilon$. Hence $x'_n \to x$.
Now consider a convergent series $\sum_n x_n$. If we let $s_n = x_1+...+x_n$, then we have $s_n \to s$.
Now consider the series $\sum_n x'_n$, where for $n \ge M$, then $x'_n = x_n$.
Let  $s'_n = x'_1+...+x'_n$. Note that for $n \ge M$, we have $s'_n -s'_{M-1} = x'_M+...x'_n = x_M+...x_n$, and so $s'_n -s'_{M-1} = s_n -s_{M-1} \to s -s_{M-1}$.
Hence $s'_n \to (s -s_{M-1}+s'_{M-1})$.
