Show compactness of an operator and calculate its SVD 
Consider

$$T\colon\ell^2\to\ell^2, (s_1,s_2,\ldots)\mapsto (s_2,s_3,\ldots)$$
$$S\colon\ell^2\to\ell^2, (s_n)\mapsto (s_n/n)$$
$$R:=TS.$$

1) Show that $R$ is a compact operator.
2) Calculate the Singular Value Decomposition of $R$.


EDIT: My result concerning 2.)
The singular value decomposition 
$$
\left\{(\sigma_j),(u_j),(v_j)\right\}
$$
of $R$ is given by
$$
\sigma_j=\frac{1}{j+1},~~~~~v_j=e_j, j\geq 1,~~~~~u_j=e_j, j\geq 2,
$$
where $\left\{u_j\right\}$ is an Orthonormalsystem of $\ell^2$ and $\left\{v_j\right\}$ is an Orthonormalsystem of $\ell^2$.
Is that right? Is that the SVD of $R$?
 A: No, this isn't correct.
In your fourth paragraph, you chose an arbitrary sequence $(s_n)$ from $\ell_2$. It need not be bounded (limited). Your argument just shows that each $s_n$ is in $\ell_2$. The constant $C$ you found depends of the particular $s_n$ that you are looking at. This error is easily remedied: you simply assume at the start that $(x_n)$ is a bounded   sequence in $\ell_2$.
A more egregious error occurs later. The vector $u$ that you constructed is not a subsequence of $(R(s_n))$; it's just an element of $\ell_2$. So, it seems you tried to identify a candidate for a subsequential limit point; but, I don't think you can show that some subsequence of $(R(s_n))$ converges $u$. You need to be more careful with your diagonalization.
But, if you want to use  a diagonalization argument, you can do the following:
Assume $(s_n)$ is a bounded sequence in $\ell_2$. If we can show that $(R(s_n))$ has a convergent subsequence, it will follow that $R$ is compact.
Towards this end, set $y_n=R(s_n)$. Then "diagonalize" $(y_n)$ as follows:
Choose a subsequence $(y^1_n)$ of $(y_n)$ so that its first coordinates form a convergent sequence in $\Bbb R$: $\lim\limits_{n\rightarrow\infty} y^1_n(1)=z_1$ for some number $z_1$.
Then choose a subsequence $(y^2_n)$ of $(y^1_n)$ so that its second coordinates form a convergent sequence in $\Bbb R$: $\lim\limits_{n\rightarrow\infty} y^2_n(2)=z_2$ for some number $z_2$.
Continue...
Now set $x_n=y_n^n$. Then $(x_n)$ is a subsequence of $(y_n)$ such that for each $i$ we have $\lim\limits_{n\rightarrow\infty} x_n(i)=z_i$.
Now show that $z\in \ell_2$ and that $\Vert x_n-z\Vert\rightarrow 0$.
(Alternatively, you could just ignore $z$ and show that $(x_n)$ is Cauchy in $\ell_2$.)


I'll remark that the argument can be simplified a tad by noting that, since $T$ is bounded and since the composition of a bounded operator with a compact operator is compact, it suffices to show that $S$ is compact.
Showing that $S$ is compact can be done without diagonalizing explicitly by, for instance, showing that $S(B(\ell_1))$ is totally bounded. Or, by showing that $S$ is the norm limit of finite rank operators (see, this post, for example).
This post will also prove helpful.
