Let $D$ be a dense subset of a banach space $X$. Show that any $x$ can be written as a sum of elements of $D$ with a certain conditon. 
Let $D$ be an everywhere dense subset of a Banach space $B$ with norm $\|\cdot\|$. Show that any $x\in B$ can be written as the sum of series $$x=\sum_{k=1}^\infty x_k\,,$$ where $x_k\in D$ and $\|x_k\|\le 3\cdot 2^{-k}\|x\|$ for every $k\ge 1$.

I am not really sure how to do this problem. It feels like i have to define $x_n$ inductively but I could not figure out the specifics. Any hints or solutions appreciated. Perhaps the trick $\sum y_n-y_{n-1}=y_n 
 $ could come in handy, but again I could not figure it out as I did not know how to force the bound onto $y_n-y_{n-1}$
 A: Hint: you want $x_1$ to be close to $x$, which can be done.  And then you want $x_2$ close to $x - x_1$...
EDIT:
Actually you can get it with $\|x_k\| \le  2^{1-k} \|x\|$.
We may assume $x \ne 0$, otherwise take all $x_k = 0$.
Lemma: Given $x \ne 0$ and $\epsilon > 0$, there is $y \in D$ with $\|y\| \le \|x\|$ and $0 < \|y - x\| < \epsilon$.
Proof: take $y$ in the intersection of $D$ with the open ball of radius $\epsilon/2$ about $r x$ where $r = 1 - \epsilon/(2 \|x\|)$.
I will inductively choose $x_k$ such that
$$0 < \left \|x - \sum_{j=1}^{k} x_j \right\| \le 2^{-k} \|x\|\ \text{and}\ \|x_k\| \le 2^{1-k} \|x\| $$
This can be done using the above lemma, rewriting the first
inequality above as $$0 < \left\|x_k - \left(x - \sum_{j=1}^{k-1} x_j\right)\right\| \le 2^{-k} \|x\| $$
and noticing that $$2^{1-k} \|x| \ge \|x - \sum_{j=1}^{k-1} x_j\|$$
A: So, we can play around with our representation of $x$ in the following way:
\begin{align*}
 \left\Vert x - \sum_{k = 1}^{\infty} x_k \right \Vert &= \left\Vert x \sum_{k = 1}^{\infty} 2^{-k} - \sum_{k = 1}^{\infty} x_k \right \Vert \\
&=  \left\Vert \sum_{k = 1}^{\infty} 2^{-k} x - \sum_{k = 1}^{\infty} x_k \right \Vert \\
&=  \left\Vert \sum_{k = 1}^{\infty}(2^{-k}x - x_k) \right \Vert \\
&\leq \sum_{k=1}^{\infty} \left\Vert 2^{-k}x - x_k  \right \Vert
\end{align*}
Now, given any $\epsilon,$ since we have a dense set $D$, we can inductively choose $x_k$ such that
$$\left\Vert x_k - 2^{-k} x \right\Vert < \frac{\epsilon}{2^{-k}}$$
since $2^{-k} x \in D.$ Now, returning to the chain previously constructed, we see:
$$\left\Vert x - \sum_{k = 1}^{\infty} x_k \right \Vert \leq  \sum_{k=1}^{\infty}\frac{\epsilon}{2^{-k}} = \epsilon $$
Therefore, we have our representation. Moreover, we see for any $k$:
\begin{align*}
\left\Vert x_k \right\Vert &= \left\Vert x_k - 2^{-k}x + 2^{-k}x \right\Vert \\
&\leq \left\Vert x_k - 2^{-k}x \right\Vert + \left\Vert 2^{-k}x \right\Vert \\
&< \frac{\epsilon}{2^{-k}} + 2^{-k} \left\Vert x \right\Vert \\
&\leq 3 \cdot 2^{-k} \left\Vert x \right\Vert
\end{align*}
Now, I should include a disclaimer that I'm not really sure about the relevance of the 3. It could be your professor attempted a sum of three small values, or applied another trick that resulted in the necessity of 3. But I don't find that the 3 is necessary and you can reduce the answer to something like:
$$\left\Vert x_k \right\Vert \leq 2^{1 - k} \left\Vert x \right\Vert$$
