Is proving that a limit exist equivalent to show that its value is real (finite)?

Question

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.

Answer 1

First of all, I upvoted; nice work, nicely shown.

I see some areas where your analysis needs improvement:

(1)
You should have expressed
$$ \sum_{n=m}^{\infty} a_n \text{ as } \sum_{n=m}^{m+k-1} a_n + \sum_{n=m+k}^{\infty} a_n. $$

This is different from what you wrote.

(2)
Continuing with your approach here (which I like), with the above correction in place,
the first term on the RHS : $\sum_{n=m}^{m+k-1} a_n$
is a summation of a fixed number of terms (and therefore finite), since $m$ and $k$ are (I assume) fixed numbers.

Therefore, employing your approach, I would have written that
$S = \sum_{n=m}^{m+k-1} a_n$, with $S$ independent of $N$,
and then written $T_N = \sum_{n=m+k}^{N} a_n. $

Then, for simplicity of notation, I would have written:
Let $T = \lim_{N \to \infty} T_N.$

(3)
Then the problem would be reduced to showing that $T$ is finite (rather than infinite) if and only if $(T + S)$ is finite.

This is the whole point of the problem, and this is where you want your intuition to expand. The above if and only if assertion should be straight forward to demonstrate using the $\epsilon, \delta$ definition from your class re infinite summation.

This is because it is clear that $\sum_{n=m}^N a_n = S + T_N.$

Can you take it from here?