A problem on sequences and series Let $ (x_n) $ be a sequence of real numbers such that $ \lim x_n =0 $.
Prove that there exists a subsequence $(x_{n_k} )$of $ (x_n) $ such that $ \sum_{k=0}^\infty 2^{k}x_{n_k}$ coverges and $\sum_{k=0}^\infty 2^{k}x_{n_k}$ converges and $\sum_{k=0}^\infty| 2^{k}x_{n_k}|\leq 1$
Didn't have any idea on how to prove or approach it so any help would be appreciated.

$\exists k_1\in N$ such that $\displaystyle |x_{m}|<\frac{1}{2^2},\forall m\ge k_1$ choose $n_1=k_1$
$\exists k_2\in N$ such that $\displaystyle |x_{m}|<\frac{1}{2^4},\forall m\ge k_2$ ,its clear that $k_2\ge k_1$choose $n_2=k_2+1$ so $n_2>n_1$
In this way choose the sub sequence(by induction) $\{x_{n_k}\}$ with $\displaystyle |x_{n_k}|<\frac{1}{2^{2k}}$
So we have $\displaystyle |x_{n_k}|<\frac{1}{2^{2k}}\Rightarrow 2^k|x_{n_k}|<\frac{1}{2^{k}}$
So we have $\displaystyle \sum_{k=1}^{\infty}x_{n_k}$ a convergingseries (by comparison test) and ,
$\displaystyle \sum_{k=1}^{\infty}x_{n_k}\le \sum_{k=1}^{\infty}\frac{1}{2^{k}}=1$
