# If $\sum|a_n|$ diverges and $\sum a_n$ converges then $\sum a_n^+$ diverges

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.

Proof:

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 $$a_n^+= \max{\{a_n, 0\} }$$] .

• 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. – 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 $$\displaystyle \sum_{n=0}^{+\infty} \vert a_n \vert = +\infty$$.

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