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I'm trying to summarise my understanding of infinite sequences, series, and their relationships with respect to convergence at the fundamental level. Here is what I know. How much of this is correct?

First off, here's a table of the notations that I use, and their corresponding meaning.

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My understanding is that:

  • The sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ converges if $$\lim\limits_{n\to\infty}a_n=L_{a}.$$

  • The infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges if its sequence of partial sums, $\lbrace s_n \rbrace _{n=0}^{\infty}$, has a limit, i.e.$$\lim\limits_{n\to\infty}s_n=L_{s}.$$

  • If the infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges, then the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is $0$, i.e. $$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$

  • The divergence test:

    If the the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is NOT $0$ or does not exist, then the infinite series diverges, i.e. $$\lim\limits_{n\to\infty}a_n\neq0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$$

Would seriously appreciate it if anyone could verify whether the above is accurate or incorrect in any way.

EDIT: I've modified the two limits notation that were mentioned in the comments and answers below, as well as adding the additional condition (limit does not exist or does not equal zero) for the divergence test. I appreciate all the answers/comments.

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    $\begingroup$ I would say that the notation $L_{a_n}, L_{s_n}$ looks pretty bad, these terms should not depend on $n$. $\endgroup$ – user99914 Dec 4 '17 at 5:08
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    $\begingroup$ BTW one useful equivalent of $\lim_{n\to \infty}a_n=L,$ that students often don't think of , is : For every $r>0$ the set $\{n:a_n\not \in [-r+l,r+L]\}$ is finite. $\endgroup$ – DanielWainfleet Dec 4 '17 at 5:49
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Everything is correct, though the notation $\lim\limits_{n\to\infty}a_n=L_{a_n}$ is nonstandard. Perhaps an improvement would be to write it as $\lim\limits_{n\to\infty}a_n=L_{a}.$ Normally just an $L$ will suffice but if you're working with multiple sequences (such as $a_n, b_n, c_n$) then $L_a$ is a good notation for the limit of the sequence $a$.

The reason why $L_{a_n}$ is not so good is because $n$ is just a free variable that has no importance whatsoever; if you write $a_n$ or $a_k$ it's the same thing. What matters is the sequence whose name is "$a$".

Also note that your last two statements are trivially equivalent; they are contrapositives of each other. In general "If $p$ then $q$" is stating the same thing as "If not $q$ then not $p$".

Therefore, just from knowing

$$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$

you can deduce

$not (\lim\limits_{n\to\infty}a_n=0 ) \rightarrow not (\sum\limits_{n=0}^{\infty}a_n \: converges) $

or in other words

$\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$

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  • $\begingroup$ Regarding the redundancy in my understanding, although $\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges,$ it does not mean that $\lim\limits_{n\to\infty}a_n=0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: converges,$ am I right? $\endgroup$ – user98937 Dec 4 '17 at 13:11
  • $\begingroup$ @user98937 Yes that's perfectly right. What you wrote is the converse of $\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$. Converses are sometimes true, but not in this case as you can see from $\sum_{n=1}^{\infty} \dfrac 1n$ which diverges, in spite of the fact that $\lim_{n \to \infty} \dfrac 1n=0$ $\endgroup$ – Ovi Dec 4 '17 at 15:33
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Everything you wrote looks good to me.

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