proving that bs is banach

Let's define $B_s$ as the of real valued sequences $(x_n)$, such that $sup_{N\in \mathbb N} |\sum_{k=0}^{N}{x_k}|$ is bounded, and make it a vector space considering the usual pointwise operations on sequences and scalars. Then define a norm by $||(x_n)||=sup_{N\in \mathbb N} |\sum_{k=0}^{N}{x_k}|$.

Clearly this space contains properly the set $C_s$ that consist of convergent series( for example $x_n = (-1)^n$).

I want to prove that:

i) $B_s$ is a banach space

$ii)$ $C_s$ is a closed subspace of $B_s$.

I tried by finding explicitly the limit of a Cauchy sequence, I think that if I have a Cauchy Sequence $(x_n)_n$ in $B_s$ given by $x_n=(x_{nk})_k$ , then the limit it's given by the pointwise limit ( I actually proved that the pointwise limit exist) But I can't prove convergence. Please I really need help with this problem )=

It's probably easier if you consider the space $\ell^\infty$ of all bounded (real) sequences and its subspace $c$ of convergent sequences. On $\ell^\infty$, you take the norm $\lVert y\rVert_\infty = \sup \lvert y_k\rvert$.
Then $I \colon \ell^\infty \to B_s$,
$$I(y) = (y_0, y_1 - y_0, y_2 - y_2, \dotsc)$$
• Please could you write more explicitly the map $I$ I don't understand )= – Shanks Sep 12 '13 at 0:32
• Let $x = I(y)$. Then $x_0 = y_0$, and for $k > 0$, we have $x_k = y_k - y_{k-1}$. The inverse is given by $S \colon B_s \to \ell^\infty$, with $S(x) = (x_0, x_0+x_1,\dotsc)$, or, if $y = S(x)$, then $y_k = \sum\limits_{j=0}^k x_j$. – Daniel Fischer Sep 12 '13 at 0:40