The problem goes as follows:

Let $a_n \in \mathbb{R}$, such that $\sum_{n=1}^{\infty} |{a_n}| = \infty$ and $\sum_{n=1}^{m} {a_n} \to a \in \mathbb{R}$ as $m \to \infty$. Let $a_n^+= \max{\{a_n, 0\} }$. Show that $\sum_{n=1}^{\infty} a_n^+= \infty$.

Please do check the proof and make necessary corrections.


We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $\infty$.

Now, denoting the set of entirety of negative numbers by $N= \{a_u\}_n $ and positive numbers $P= \{a_v\}_n = \{a_v^+\}_n $ We know, $a_u= -|a_u| \ \forall u \ \in N$

We can write $\sum_{n=1}^{\infty} {a_n} =\sum_{v=1}^{s} {|a_v|} -\sum_{u=1}^{t} {|a_u|} = a \ ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $\sum_{n=1}^{\infty} |{a_n}| = \infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2\sum_{v=1}^{s} {|a_v|} = \infty + a =\infty \implies$ $\sum_{n=1}^{\infty} a_n^+= \infty $ [ The negative terms can be considered to be vanished by $a_n^+= \max{\{a_n, 0\} }$] .

  • 1
    $\begingroup$ More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $\sum a_n$ and $\sum a_n^+$ both converge then $\sum a_n^-$ converges as well hence $\sum|a_n|$ converges. Now take the contraposition. $\endgroup$ – Did Dec 2 '18 at 9:17

For all $n \in \mathbb{N}$,

$$ a_n^+ = \max(a_n, 0) = \frac{a_n + \vert a_n \vert}{2}. $$

Therefore, for $N \in \mathbb{N}$ :

$$ \sum_{n=0}^{N} a_n^+ =\frac{1}{2} \sum_{n=0}^{N} a_n + \frac{1}{2} \sum_{n=0}^{N} \vert a_n \vert $$

where $\displaystyle \sum_{n=0}^{N} a_n \; \to \; a$ as $N \to +\infty$ since $\displaystyle \sum_{n=0}^{+\infty} a_n = a$ and $\displaystyle \sum_{n=0}^{N} \vert a_n \vert \to +\infty$ since $\displaystyle \sum_{n=0}^{+\infty} \vert a_n \vert = +\infty$.

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    $\begingroup$ Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof? $\endgroup$ – Martin R Dec 2 '18 at 9:18
  • $\begingroup$ can you please check the validity of the solution $\endgroup$ – Subhasis Biswas Dec 2 '18 at 9:35

(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.


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