Proving compactness of $A:l_2\to l_2\:\:Ax=(0,x_1,\frac{1}{2}x_2,...,\frac{1}{n}x_n,...)$ Problem: Show that the operator $A:l_2\to l_2\:\:Ax=(0,x_1,\frac{1}{2}x_2,...,\frac{1}{n}x_n,...)$ is compact but has no pontual spectrum.
My solution: $(0,x_1,\frac{1}{2}x_2,...,\frac{1}{n}x_n,...)=(\lambda x_1,\lambda x_2,...,\lambda x_n,...)$
Since $\lambda x_1=0\implies \lambda=0$, then the $\ker(A-\lambda)=0$.
My problem lies with compactness. 
If we consider a sequence in $l_2$ like $\{x\}_n$,that converges to some $x$. Consider $\epsilon>0$ so that $||x_m-x||<\epsilon$ for $m>N\in\mathbb{N}$. Then $||A(x_m)-A(x)||^2=||A(x_m-x)||^2=\sum_\limits{n=1}^{\infty} |\frac{1}{n}({x_m}_i-x_i)|^2<\sum_\limits{n=1}^{\infty} |\frac{1}{n^2}(\frac{\epsilon^\frac{1}{n}}{2})^n|<c\epsilon<\delta\:\:\:\delta>0$ Therefore we can find a subsequence that converges proving that $A$ is compact.
Question:
Is my proof right? If not. How should I prove compactness?
Thanks in advance!
 A: Your argument only shows continuity. An easy way to show compactness is to prove that $A$ can be written as a sum of a finite-rank operator plus an operator with arbitrarily small norm (simply by truncating). 
A: To show that the operator is compact we must show that A is linear and that the finite-rank operator
$$A_n x=(0,x_1,\frac1{2}x_2,\dots,\frac1{n}x_n, 0, \dots)$$ is bounded. 
First, we show bounded.
$$||A_n x||^2=\sum_{j=1}^n |\frac1{j}x_j|^2 \leq \frac1{1} \sum_{j=1}^n |x_j|^2 \leq ||x||^2$$
Since we have a finite rank operator that is bounded, $A_n$ is compact.
Now we want to show that the norm $||(A-A_n)x||$ goes to $0$ as $n$ goes to infinity. In other words, the limit of finite rank operators $A_n$ is A. 
$$||(A-A_n) x||^2=\sum_{j=1}^\infty |\frac1{j}x_j|^2 - \sum_{j=1}^n|\frac1{j}x_j|^2$$
$$=\sum_{j=n+1}^\infty|\frac1{j}x_j|^2 \leq \frac1{(n+1)^2}\sum_{j=n+1}^\infty|x_j|^2 = \frac1{(n+1)^2} ||x||$$ 
So that as we take $n\rightarrow \infty$, $||(A-A_n) x||^2 \rightarrow 0$
Finally to show that the operator has no point spectrum. 
To be in the point spectrum, we must have that $$R_\lambda = (A-\lambda I)^{-1}$$ does not exist. 
First take $\lambda \neq 0$
We have that $$T_\lambda x = 0 \rightarrow (A-\lambda I)x= Ax-\lambda x =0$$
$$\rightarrow (0,\frac1{2}x_1,...,\frac1{n},...)-\lambda (x_1,x_2,...,x_n,...)=0$$ leads to the recurrence relation of 
$$x_1=0$$ and $$x_{j}=\frac1{j-1} \frac1{\lambda}x_{n-1}$$ for $j\geq 2$. 
This only holds for $x=(x_1,x_2,...)=0$. Thus, the nullspace of $T_\lambda =\{0\}$ and so $T_\lambda^{-1}=R_\lambda$ exists and $\lambda$ is not in the point spectrum.
Now take, $\lambda =0$. This means that $T_\lambda = Ax = 0$ implies that $x=0$ again so that $\lambda$ is not in the point spectrum again for the same reason as before.
Therefore, the point spectrum is empty.
