I'm studying Tao analysis I. My question arises from proving results using limit law, this is an example from proposition 7.2.14(c):
c) Let $\sum\limits_{n=m}^{\infty}a_n$ be series of real numbers, and let $k\geq 0$ be an integer. If one of the two series $\sum\limits_{n=m}^{\infty}a_n$ and $\sum\limits_{n=m+k}^{\infty}a_n$ are convergent, then the other one is also ,and we have the following identity $$\sum\limits_{n=m}^{\infty}a_n=\sum\limits_{n=m}^{m+k-1}a_n +\sum\limits_{n=m+k}^{\infty}a_n$$
My attempt to prove: Let $S_N=\sum\limits_{n=m}^{N}a_n$ and $T_N=\sum\limits_{n=m+k}^{N}a_n$, then we have $S_N=\sum\limits_{n=m}^{m+k-1}a_n+T_N$ for all $N\geq m+k$, (the statement also hold when $N<m+k$ with $T_N=0$ and $S_N$ has redundant zero terms after index $N$ ) , taking the limit as $N\to \infty$, we have $$\lim_{N\to\infty}S_N=\lim_{N\to\infty}\sum\limits_{n=m}^{m+k-1}+\lim_{N\to\infty}T_N$$ $$=\sum\limits_{n=m}^{m+k-1}+\lim_{N\to\infty}T_N,$$ since the finite sum is independent on $N$.
Now, assume $\sum\limits_{n=m}^{\infty}a_n$ converges to $L$ , then $\lim_{N\to\infty}S_N$ exists and equals $L$, and let $\sum\limits_{n=m}^{m+k-1}=M$, since finite sums are convergent, my question is can we use the previous two result to conclude that $\lim_{N\to\infty}T_N$ exists and equals $L-M$.
Or should I prove that $S_N$ is a Cauchy sequence if and only if $T_N$ is? Again, I'm not looking for a solution or a proof verification, my question as the title says: is proving the existence of a limit equivalent to showing that it's value is finite or not?
In a more logical terms is the following $equivalence$ statement true : limit exists $\longleftrightarrow$ limit's value $\in \mathbb{R}$.
If yes, why can't we assume that limits exists, then try to compute its value and if it's real then conclude that it exists, for instance in evaluating $\lim\limits_{n\to\infty}x^n$ and equals $L$, then $xL=\lim\limits_{n\to\infty}x^{n+1}=L$ , then we have $(x-1)L=0$. Since $x=1$ for every real $x$ is absurd, we conclude that $L=\lim\limits_{n\to\infty}x^n=0$ when $x\neq 1$ . However we know that the above reasoning is false since the limit doesn't exist in the first place.