Is it possible to find sequence $ a_k $ such that $ \sum_{k=1}^{\infty}a_{k},\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k} $ Is it possible to find sequence $ a_k $ such that $ \sum_{k=1}^{\infty}a_{k},\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k} $ both converge conditionally ?
I think that its possible, I tried to find an example (like a sequence $ a_k $ that is positive when $ k=1,0 $ module 3, and negative when $ k=2 $ module 3. But didnt succeed. Is it true at all?
Thanks in advance
 A: It's even easy to characterize such sequences. $A=\sum_{k=1}^\infty a_k$ and $A_\text{alt}=\sum_{k=1}^\infty(-1)^k a_k$ both converge if and only if $A+A_\text{alt}=2A_\text{even}=2\sum_{k=1}^\infty a_{2k}$ and $A-A_\text{alt}=2A_\text{odd}=2\sum_{k=1}^\infty a_{2k-1}$ both converge. Hence, $A$ and $A_\text{alt}$ both converge conditionally (i.e. not absolutely) if and only if $\sum_{k=1}^\infty|a_k|$ diverges but both $A_\text{even}$ and $A_\text{odd}$ converge (and then at least one of the last two must be conditionally convergent).
In particular, we may take any conditionally convergent $\sum_{k=1}^\infty a_{2k}$ and put $a_{2k-1}=0$.
A: 
Is it possible to find sequence $a_k$ such that $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty (−1)^ka_k$ both converge conditionally ?

Yes.  Let $a_k = \frac{\sin(k)}{k}$.  Then, both the series $\sum_{k=1}^\infty \frac{\sin(k)}{k}$ and $\sum_{k=1}^\infty \frac{(-1)^k\sin(k)}{k}$ are conditionally convergent.
Dirichlet's Test guarantees that both series converge.  To show that the convergence is only conditional, it is sufficient to show that $\sum_{k=1}\frac{|\sin(k)|}{k}$ diverges See here for proofs.
A: in complex analysis at least it is simple:
https://math.stackexchange.com/a/118734/427318
The series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges everywhere on the boundary of the unit disk other than z=1, but thus not absolutely.
now take z=i, or -i, or indeed any complex number of magnitude 1, that is not 1/-1  to get your series
