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

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 $$N with $$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.

• The point where you take the limit has a slight problem. Since the limit of $S_N$ exists you know that the limit of the RHS exists but you can't break it into two limits since you don't yet about $T_N$. Since you don't know if the limit of $T_N$ exists, isolate in one side of the equation, argue that every single limit in the other side exists, so the sum exists and deduce the existence of the limit of $T_N$.
– cgss
Aug 19, 2020 at 0:20
• Recall what the definition of a limit is. We say a sequence $(x_n)$ converges to $L\in \mathbb{R}$ if.... Therefore, in the definition, you require the limit it be a real (finite) number if a sequence converges. Also in your example with $x^n$, the only limits will be $0$ if $|x| <1$ or $1$ if $x = 1$, otherwise the sequence does not converge. In either case $(x-1)L = 0$ since either $L=0$ or $x=1$. Aug 19, 2020 at 0:24
• @cgss,do you mean to write $T_N=S_N-\sum\limits_{n=m}^{m+k-1}$, and since the limits, at the RHS exists, then the limit of $T_N$ exists. Aug 19, 2020 at 8:34
• @Dayton, I know that the proof of the limit of $x^n$, is actually a proof by contradiction since the assumption leads to absurdity and hence should be wrong, however I'm asking is the equivalence between existence and evaluation of limit to a finite real number true or not, if yes why can't we use the mentioned schema of assumption then evaluating then reconfirming the assumption. Aug 19, 2020 at 8:52
• @Kareem Yes. It's a technicality of course but sometimes even those are important.
– cgss
Aug 19, 2020 at 10:18

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?

• @KareemTaha Your comment is wrong. $T_N = \sum_{m+k}^N a_n.$ Aug 19, 2020 at 8:11
• Ah. Thanks, re-corrected it in the question. Aug 19, 2020 at 8:13
• Please add which series you assumed to be convergent and which one we are trying to prove it's convergence from the assumption. Aug 19, 2020 at 8:15
• @KareemTaha By the way, just because an infinite series is finite does not mean that it is convergent. Consider the series $a_n = (-1)^n.$ As $n \to \infty,$ the summation will oscillate back and forth from -1 to 0. It is considerations like this that make me favor an $\epsilon, \delta$ demonstration, rather than analysis that is founded on the idea that the infinite summation is finite. Aug 19, 2020 at 8:42
• @KareemTaha If you still need the question re equivalence addressed then I would ask you to :(1) Overhaul your query to remove any mistakes (for example, your use of $S_N,$ use my answer as a guide). (2) Try to complete the problem (perhaps with $\epsilon, \delta$) and (again) edit your query to show all of your (new) work. (3) If you have a question that involves something like (for example) Cauchy sequences, include a definition of (for example) Cauchy sequences. This makes it easier to respond to your questions. Aug 19, 2020 at 8:48