$\sum_{n\in\mathbb{N}}a_n$ conditionally convergent$\Rightarrow \nexists X\subseteq\mathbb{N}$ infinite s.t. $\sum_{n\in X}a_n$ absolutely convergent I'm wondering about the following:
if $\sum_{n\in\mathbb{N}}a_n$ is a conditionally (but not absolutely) convergent series then there does not exists $X$ infinite subset of $\mathbb{N}$ such that $\sum_{n\in X}a_n$ is absolutely convergent;
is it true in general? Intuitively I think the answer is yes, but I'd like to see a proof/counterexample.
Best regards,
lorenzo.
 A: Consider the alternating harmonic series 
$$
\sum_{k = 1}^\infty \frac{(-1)^{k-1}}{k} = \ln2;
$$
it converges however not absolutely. However, with $X = \{$square integers$\}$ the series converges absolutely. This is a counterexample.
A: The opposite is true.  If $\sum a_n$ converges conditionally but not absolutely, then there exists infinite $X$ such that $\sum_{n \in X} a_n$ converges absolutely.  
To see this:  sequence $a_n$ goes to zero, but is not eventually zero.  So we may choose $n_1$ such that $0<|a_{n_1}| < 1/2$, then choose $n_2 > n_1$ such that $0<|a_{n_2}| < 1/2^2$, then choose $n_3 > n_2$ such that $0<|a_{n_3}| < 1/2^3$, and so on.  We recursively choose the sequence $n_1 < n_2 < n_2 < \dots$ such that $|0<|a_{n_k}| < 1/2^k$ for all $k$.  Then the set $X = \{n_1, n_2, n_3,\dots\}$ is the set required.
A: Recall that $\sum_{n\geq1}(-1)^{n}/n=-\log2$, but $\sum_{n\geq1}1/n$
is divergent. Moreover, $\sum_{n\geq1}1/n^{2}$ is absolutely convergent.
We can embed the absolutely convergent sequence in the midst of the
conditionally convergent sequence by considering the series $\sum_{n\geq1}f(n)$
where
$$
f(n)=\begin{cases}
\frac{(-1)^{n/2}}{n/2} & \text{if }n\text{ is even};\\
1/\left\lceil n/2\right\rceil ^{2} & \text{if }n\text{ is odd}.
\end{cases}
$$
A: No. Consider
$$
a_n=\begin{cases}
0 & \text{when}\ n \text{odd}\\
\frac{(-1)^{n/2}}{n/2} & \text{when}\ n \text{even}
\end{cases}
$$
That's essentially just the alternating harmomic series, and as such well-known conditionally convergent, but it's also obvious that the odd natural numbers form an infinite subset of indices $X$, such that
$$
\sum_{n\in X} a_n
$$
is absolutely convergent.
