Question the convergence of the following: Question the convergence of the following:


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*The sequence $\{x_n\}_{n=1}^{\infty}$ in $l_1$, 
where $x_n=(\frac{1}{2^n},\frac{1}{2^n},...,\frac{1}{2^n},0,0,...)$
(here the first $n$-places have the value $\frac{1}{2^n}$)

*The sequence  $\{x_n\}_{n=1}^{\infty}$ in $l_2$, 
where $x_n=(n\sin\frac{1}{n},n\sin\frac{1}{n},...,n\sin\frac{1}{n},0,0,...)$
(here the first $n$-places have the value $n\sin\frac{1}{n}$)

*The series $\sum_{n=1}^{\infty}x_n$ in $l_2$ where $x_n=(\frac{1}{n},\frac{1}{n^2},...,\frac{1}{n^n},0,...)$
First off, I must say, this does seem a little ambiguous. But I have this information from class, which might help me, and others as to what is questioned here. I must do this type of question as it is on the exam.
I have that $x_n\to^{l_r}x$ (converges to $x$ in $l_r$) if
$\|x_n-x\|_{l_r} \to 0$ and $n \to \infty$, meaning that if $x_n=(x_n^1,x_n^2,x_n^3...)$ and $x = (x^1,x^2,....)$ then 
$$(\sum_{k=1}^{\infty}|x_n^k-x^k|^r)^{\frac{1}{r}}\to^{n\to \infty} 0.$$
So I gather that I have to find an $x \in l_1$(in the first example) such that $$\sum_{k=1}^{\infty}|x_n^k-x^k| \to^{n\to \infty} 0 \iff\lim_{n\to \infty}|x_n^k-x^k|=0$$
I really need help with this. Much appreciation. For the first one I would think that $(0,0,0,...) \in l_1$ and also $\lim_{n\to \infty}|\frac{1}{2^n}|=0.$

*

*$\Vert x_{n}\Vert_{1}=n\frac{1}{2^{n}}\rightarrow0$. Therefore, $x_{n}\rightarrow 0$ in $l_1$.

*Note that $n\sin\frac{1}{n}\rightarrow1$. Therefore, $x_{n}\rightarrow e=(1,1,\ldots)$ pointwise. Since $\Vert e\Vert_{2}=\infty$, what can you conclude?

*If $\sum_{i=1}^{n}x_{i}$ is convergent, then so too is $\sum_{i=1}^{n}\frac{1}{i}$. But we know this not to be the case.

