If a series converges, does +1 to its upper index also converge to the same limit point? Almost certainly yes, but how do you prove it?

Question

I encountered a proof that assumed the following is true, without proving it or referencing it anywhere so this got me wondering how they knew it was true.

Suppose the series $\lim_{k \rightarrow \infty}\sum_{n=0}^{k}||a_n||$ converges to some number $L$ where each $||a_n||$ is non-negative.

Then is it also true that

$\lim_{k \rightarrow \infty}\sum_{n=0}^{k+1}||a_n|| = L$?

Probably, but I've never been good at anything involving indices and sub-sequences, so I have no idea how to go about this.

I know

\begin{align*} \lim_{k \rightarrow \infty} || \sum_{n=1}^{k+1} - L|| \leq \lim_{k \rightarrow \infty}(|| \sum_{n=1}^{m} - L|| + ||a_{k+1}||) \\ = 0 + \lim_{k \rightarrow \infty}||a_{k+1}|| \end{align*}

Then, from a surprisingly not-proven "theorem" in one of Rudin's principles of analysis books (3.23), I "know" the terms of $||a_k|| \rightarrow 0.$

I'm unsure at the moment what I should do about this extra shifted index. If I re-write the index like $l = k+1$ then I get some other thing I don't know how to prove in the lower limit. The sequence $||a_{k+1}||$ is not necessarily monotonic so I can't bound it above or below by $||a_k||$ either.

Throwing a quick guess... if a sequence converges, then all of its subsequences converge (to the same limit as the parent sequence). See https://math.stackexchange.com/questions/213285/prove-if-a-sequence-converges-then-every-subsequence-converges-to-the-same-lim for example. So adding more more term could be seen as an application of this.
Also, from the definition of converge, for some $n \geq N$ $|a_n - l| < \epsilon$ for any $\epsilon > 0$. So an extra term wouldn't change that since $k+1 > N$. — ning, Sep 06 '23 at 22:18
It sounds like you can simply the question (to something that's still confusing to you in this moment): if you have any sequence $(s_k)$ such that $\lim_{k\to\infty} s_k = L$, is it also true that $\lim_{k\to\infty} s_{k+1} = L$? The answer is yes. The justification depends on how you're being taught about sequences and what tools you've been provided. — Greg Martin, Sep 06 '23 at 22:21
Assume the very basics, like the early chapters of Rudin's principles of mathematical analysis, though don't assume I am an expert on topology. What ning said seems plausible to work from the definition of a convergent sequence, perhaps I can stare at the long enough and make sense of it. — CheeseBlues, Sep 06 '23 at 22:22
Yes, I would argue that this question is a duplicate of the linked one. The fact that it's a series is a complete ruse; simply consider it as a sequence $s_k=\sum_{n=0}^k |a_n|$. Then the subsequence in question is $s_{k_i}=s_{i+1}$. Of course there are more elementary arguments, as ning provides in their comment, but the general result is also relatively straightforward and good to know. — Eric Nathan Stucky, Sep 06 '23 at 23:06

If a series converges, does +1 to its upper index also converge to the same limit point? Almost certainly yes, but how do you prove it?

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