Totally boundedness of the set $A= \{(x_n)_{n=0}^{\infty} : \sum_{n=0}^{\infty}n \cdot \lvert x_n \rvert \le 5 \}$ in sequence spaces. $A= \{(x_n)_{n=0}^{\infty} : \sum_{n=0}^{\infty}n \cdot \lvert x_n \rvert \le 5 \}$. Is this set totally bounded in the spaces: 
(a) $X=l_1$, $d(x,y)= \sum_{n=0}^{\infty} \rvert x_n -y_n \rvert $ 
(b) $X=l_2$, $d(x,y)= (\sum_{n=0}^{\infty} (\rvert x_n -y_n \rvert)^2)^{1/2} $ 
(c) $X= \omega $, $d(x,y)= \sum_{n=0}^{\infty} 2^{-n}\frac{\rvert x_n -y_n \rvert}{1+\rvert x_n -y_n \rvert}$, where $\omega = \mathbb{R}^{\mathbb{N}}$. 
I do not have any idea and experience in these calculations. Can you help me? Thanks for any help.
 A: EDIT: I had missed the fact that indexing of sequences starts from $0$ instead of $1$. It is obvious that $A$ is unbounded, so it cannot be totally bounded. 
If indexing starts from $1$ instead of $0$ then total boundedness is true in all three cases: 
Hints for 1) and 2): Take any sequence $(x_n^{(j)})$ in $A$. Use a diagonal procedure to extract a subsequence $(x_n^{(j_k)})$ such that $x_n=\lim_{k \to \infty} x_n^{(j_k)}$ exists for each $n$. Then use Fatou's Lemma to show that $(x_n) \in A$. [ It is important to note that $\sum n|x_n| <\infty$ implies that $(x_n) \in l_1$ and  $(x_n) \in l_2$]. Since every sequence in $A$ has a convergent subsequence it follows that $A$ is totally bounded. 
For 3) note that $A \subset [-5,5]^{\mathbb N}$. This product is compact by Tychonoff Theorem. [ The topology induced by your metric is same  as the product topology). Hence $A$ is subset of a compact metric space , hence totally bounded. 
A: $a).\ $To show total boundedness, for an arbitrary $\epsilon>0,$ you need to find an $\epsilon$- net that captures each element of $A$. To disprove total boundedness, you need to find an $\epsilon>0$ for which the definition fails. If we take $\epsilon=1/2$ and consider the infinite collection of sequences
$(x_n^1)=(1,0,\cdots,0,\cdots ),\ (x_n^2)=(2,0,\cdots,0,\cdots );\cdots ,(x_n^k)=(k,0,\cdots,0,\cdots )$, 
all of which are in $A$ because $\sum^{\infty}_{n=0}nx_n^k=0\le 5$, we see that the distance between any two of them is greater than $1/2$ so there is no finite $\epsilon$-net that contains them. Hence $A$ is not totally bounded.
$b).\ $  and $c).$ are handled similarly
Remark:
It may be interesting to change the exercise a bit to see better how these analyses work. Take 
$A= \{(x_n)_{n=0}^{\infty} : \sum_{n=0}^{\infty}(n+1) \cdot \lvert x_n \rvert \le 5 \}$ and do $a)$. Here is an idea:
For convenience of the argument, take $\epsilon=1.$  Now, 
if $(x_n)\in A,$ its first component $x_1$ must satisfy $-5\le x_1\le 5$. So, we can take the balls of radius $1/2$ about the sequences whose first components are  $4.75,4.5,4,3.5,3,,2.5,2,1.5,1,.5,-.5,-1,-1.5,-2,-2,5,-3,-3.5,-4,-4.5,-4.75$ and zero in every other position. There are finitely many and they are all in $\ell_1$. 
The second component of $(x_n)$ must satisfy $-5/2\le x_2\le 5/2$ so we repeat the above procedure to find finitely many balls centered at sequences whose first component is one of the above and whose second component is obtained as in the first step, but using the interval $-5/2\le x_2\le 5/2$. So now we take all possible combinations of the two components we have just defined, to get  a finite number of $\ell_1$ sequences and so a finite number of balls of radius $1/2$ centered at each. 
We continue in this way, until at the eleventh step, we have that $-5/12\le x_5\le 5/12$ and at this point we notice that if our sequence is to be in $A$, then all its components including and after this one must be at a distance from zero less than $1$.
This means that we can take all the balls of radius $1/2$ centered at the sequences we have previously constructed in steps $1$ to $10$, with zero component for $n\ge 11$. By construction, the number of such balls is finite and each member of $A$ is in one of them.
It's easy to see that this procedure will work for any $\epsilon>0$ so we have in this case that $A$ is totally bounded. 
A: The set is not totally bounded in any of the cases.
Note that $B=\mathbb{R} \times \{0\} \times \{0\} \times \cdots \subset A$.
Since $B$ is unbounded is cannot be totally bounded.
To illustrate, take the sequence $x_n = (n,0,0,...) \in A$. This has no convergent subsequence.
