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I am self-learning real analysis from Understanding Analysis by Stephen Abbot. I'd like to ask if I have deduced the correct conclusions for the below assertions about a subvergent (invented definition) series.

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Definition. Let's say that a series subverges if the sequence of partial sums contains a subsequence that converges.

Consider this (invented) definition for a moment, and then decide which of the following statements are valid propositions about subvergent series:

(a) If $(a_n)$ is bounded, then $\sum a_n$ subverges.

(b) All convergent series are subvergent.

(c) If $\sum \absval{a_n}$ subverges, then $\sum a_n$ subverges as well.

(d) If $\sum a_n$ subverges, then $(a_n)$ has a convergent subsequence.

Proof. (a) This proposition is false. As a counterexample, consider the sequence $(a_n):=1$. The sequence of partial sums is $s_1 = 1, s_2 = 2, s_3 = 3, \ldots, s_n = n,\ldots$. No subsequence of $(s_n)$ converges. So, $\sum {a_n}$ is not subvergent.

(b) Since the series is convergent, the sequence of the partial sums converges and therefore any subsequence of partial sums also converges to the same limit. Thus, all convergent series are subvergent.

(c) I think this proposition is true. Let $(s_n)$ be the sequence of partial sums of the absolute values and $(t_n)$ be the sequence of partial sums of the series $\sum a_n$.

By definition of subvergence, there is some subsequence $(s_{f(n)})$ of $(s_n)$ that converges. Without loss of generality, assume $(s_{2n})$ is one such convergent subsequence. Then, there exists a $N \in \mathbf{N}$ such that, \begin{align*} \absval{\absval{a_{2m+2}} + \absval{a_{2m + 4}} + \ldots + \absval{a_{2n}}} < \epsilon \end{align*}

for all $n > m \ge N$.

Using this fact, we can write a nice inequality for the subsequence $(t_{2n})$. \begin{align*} \absval{t_{2n} - t_{2m}} &= \absval{a_{2m+2} + a_{2m+4} + \ldots + a_{2n}}\\ &\le \absval{a_{2m+2}} + \absval{a_{2m+4}} + \ldots + \absval{a_{2n}}\\ &\le \absval{\absval{a_{2m+2}} + \absval{a_{2m+4}} + \ldots + \absval{a_{2n}}}\\ &< \epsilon \end{align*}

for all $n \ge N$.

As the above holds true for all subsequences $(s_{f(n)})$ where $f(n):\mathbf{N} \to \mathbf{N}$ is a bijection, $\sum a_n$ is subvergent.

(d) I can't think of a counterexample for this.

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    $\begingroup$ slader.com/discussion/question/… --> the answer for (d) is (1,-1,2,-2,3,-3,...,n,-n,...) $\endgroup$ – BCLC Jan 1 at 13:35
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    $\begingroup$ slader.com/discussion/question/… --> i notice you have the same answer in (b) as in here, and I find both (the same) answers to be weird. just use that every sequence is a subsequence of itself (unless there's this weird definition that subsequence excludes the original sequence. but in this case, we might just exclude every vector space as a subspace of itself. or every set as a subset of itself. idk) $\endgroup$ – BCLC Jan 1 at 13:40
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    $\begingroup$ are you sure about your (c)? slader has a different version. in slader, the assumption is $\sum | \cdot |$ instead of just $ | \cdot |$ $\endgroup$ – BCLC Jan 1 at 13:50
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    $\begingroup$ (c) I made a typo in the question. It's $\sum \lvert \cdot \rvert$. $\endgroup$ – Quasar Jan 1 at 13:52
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  • For a) your proof is ok
  • For b), ok too
  • For c), I would have written :

Let's $a_n^+=\max \{0, a_n\}$ and $a_n^- = \max \{0, -a_n\}$ for all $n$.

Then for all $n$, $|a_n|=a_n^+ + a_n^-$ and $a_n = a_n^+ - a_n^-$.

Since $\sum |a_n|$ is subvergent, and $0\leqslant a_n^+ \leqslant |a_n|$ and $0\leqslant a_n^- \leqslant |a_n|$, we have that $\sum a_n^+$ and $\sum a_n^-$ are subvergent, so the sum $\sum a_n$ is subvergent.

(The fact that if $\sum u_n$ converges with $(u_n)$ positive, then for all $(v_n)$ positive such that $\forall n,v_n\leqslant u_n$ subverges would deserve a proof, but it's not that difficult)

  • For d) I define $(a_n)$ such that for $n\geqslant 0$,

$a_{2n} = -n$ and $a_{2n+1} = n + \frac{1}{n^2}$.

Then $\sum a_n$ converges since (if we note $S_n = \sum\limits_{k=0}^n a_n$) $S_{2n+1} = \sum\limits_{k=1}^n \frac{1}{k^2}$ converges when $n\rightarrow +\infty$.

But we clearly don't have a subsequence that converges.

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  • $\begingroup$ for d is the (1,-1,2,-2,3,-3,...,n,-n,...) from comments correct? $\endgroup$ – BCLC Jan 6 at 6:40
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    $\begingroup$ Yes it is! It is even simpler than mine! $\endgroup$ – math Jan 6 at 12:18

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