If $\sum a_n = +\infty$, with $a_n ≥0$, then we can find two disjoint subseries diverging While trying to solve a problem from my measure theory class, I started to wonder about the following result:
Consider a sequence of positive terms $\left \{ a_n \right \}_{n=1}^\infty$ such that $\sum a_n = +\infty$. Is it always possible to find two disjoint subsequences $\left \{a_{n_j} \right \}_{j=1}^\infty$ and $\left \{a_{k_j} \right \}_{j=1}^\infty$ such that? $$\sum_{j=1}^\infty a_{n_j}=+\infty=\sum_{j=1}^\infty a_{k_j}$$
(Disjoint means$\ \left \{a_{n_j} : j \in \mathbb{N}\right \} \cap \left \{a_{k_j} : j \in \mathbb{N}\right \}=\emptyset$ )
Intuitively, it seems true, but I wasn't able to prove it, nor think of a counterexample. 
Given a infinite set $I\subset \mathbb{N}$, since $$\sum_{n \in I}a_n + \sum_{n \not \in I}a_n = \sum_{n=1}^\infty a_n = +\infty$$
It follows that one of the two sums on the left must be divergent. If both are, we are done. If they are not, applying the same process to the divergent sum we obtain a sequence $I_1 \subset I_2 \subset ...\subset \mathbb{N}$ such that $$\sum_{n \in I_k}a_n =+\infty \ \text{and} \ \sum_{n \not \in I_k}a_n <+\infty$$
for all $k=1,2,...$
I got this far, although I'm don't know whether this is useful at all. 
Any thoughts?
Thanks in advance!
 A: An idea is to make "slices" of your series, each slice summing to at least $1$, and define your two subsequences by looking at alternating slices.
Let $(a_n)_{n\geq 0}$ a positive sequence such that $\sum_{n=0}^\infty a_n = \infty$.
Let $N_0 = \min\{n\geq 0 \mid \sum_{n=0}^\infty a_n > 1 \}$, $N_1 = \min\{n\geq N_0 + 1 \mid \sum_{n=N_0+1}^\infty a_n > 1 \}$, and by induction define $N_j = \min\{n\geq N_{j-1} + 1 \mid \sum_{n=N_j+1}^\infty a_n > 1 \}$. (Why is it a valid definition?)
Now, you can define your two subsequences as the one with indices in $\bigcup_{j} \{N_{2j},\dots, N_{2j+1}-1\}$ and the one with indices in $\bigcup_{j} \{N_{2j+1},\dots, N_{2j+2}-1\}$.
A: Hint: No matter how many finitely many terms in the series you have assigned so far to your two sub-series, you can always find a subsequent finite partial "tail" of the terms not used so far so that the sum of the terms in the partial tail is $> \epsilon$ (and you can do this for any $\epsilon > 0$). So just fix any $\epsilon > 0$ and then assign such partial tails to each of your subsequences in an alternating fashion so they are disjoint indices, and then each sum tends to $\infty$.
A: There you got a simple proof: let $\{a_n\}$ converge to zero and such that $\sum a_n=+\infty$. Now consider $b_1\geq b_2\geq b_3 \geq \cdots$ the rearrangement of $a_n$ in decreasing order. Since $\{a_n\}$ is positive,
$$
\sum_{n=0}^\infty a_n =\sum_{n=0}^\infty b_n = \sum_{n=0}^\infty b_{2n} + \sum_{n=0}^\infty b_{2n+1}.$$
We prove that the two series in the right-hand side are divergent. It is clear that $\sum_{n=0}^\infty b_{2n}\geq \sum_{n=0}^\infty b_{2n+1}$, by construction of $b_n$. Now suppose that the latter series converges. Then,
$$
\sum_{n=1}^\infty b_{2n}\leq \sum_{n=0}^\infty b_{2n+1},
$$
which implies that 
$$\sum_{n=0}^\infty b_{2n}= b_0+\sum_{n=1}^\infty b_{2n}\leq b_0 +\sum_{n=0}^\infty b_{2n+1},$$
and by the boundedness of $a_n$ (and therefore of $b_n$), we obtain the convergence of both series, which implies that $\sum a_n<+\infty$, a contradiction. Therefore we found a decomposition of $\sum a_n$ in two disjoint divergent series. If $a_n$ does not converge to zero, just take two disjoint subsequences that do not converge towards zero and sum them up.
