Compact operator? self adjoint operator? Stirling's formula Define a pair of operators $S\colon\ell^2 \rightarrow \ell^2$ and $M\colon\ell^2 \to\ell^2$ as follows: $$S(x_1,x_2,x_3,x_4,\ldots)=(0,x_1,x_2,x_3,\ldots) $$ and $$M(x_1,x_2,x_3,x_4,\ldots)=(x_1,\frac{x_2}{2},\frac{x_3}{3},\frac{x_4}{4},\ldots).$$ Let $T=M \circ S$.
Then how to prove that $T$ has no eigenvalues? Is $T$ a compact operator? And self-adjoint?
Also, if let $T^m$ denote the composition of $T$ with itself $m$ times. Then how to show directly from the formula for $T$ that $\lim_{m\rightarrow \infty} \|T^m\|^{1/m}=0$.
I think I need to use the Stirling's formula, but I don't know what to do. Please help me. Thank you.
 A: Denote $x=(x(k))_{k\geq 1}$ a sequence in $\ell^2$. We have 
$$T^{j+1}(x)=M((T^jx(k+1))_{k\geq 1})=\left(\frac{T^j(k+1)}k\right)_{k\geq 1}$$
hence we have $(T^{j+1}(x))(k)=\frac{(T^j(x))(k+1)}k$. By induction, we get that 
$$(T^j(x))(k)=\begin{cases}
0&\mbox{ if }j\geq k\\\
\frac{x(k+j)}{\prod_{l=k}^{k+j}l}&\mbox{ if }j<k.
\end{cases}$$
We can see that $$\lVert T^j\rVert=\frac 1{\prod_{l=j+1}^{2j+1}l}=\frac{j!}{(2j+1)!}\sim\frac{\sqrt{2j\pi}(j/e)^j)}{\sqrt{2(2j-1)\pi}((2j+1)/e)^{2j+1}}=\frac{\sqrt jj^j}{\sqrt{2j-1}2^{2j+1}(j+1/2)^{2j+1}}$$
therefore 
$$\lVert T^j\rVert^{1/j}\sim \frac{j}{4(j+1/2)^2},$$
which converge to $0$. In particular, the spectral radius of $T$ is $0$, so the only possible eigenvalue of $T$ is $0$ (but we can see it's not the case because both $S$ and $M$ are injective). So $T$ has no eigenvalues. 
We can show that $M$ is compact, for example writing it has the limit of the finite rank operators $M_n$ given by $M_n(x)(k)=\begin{cases}\frac 1kx(k)&\mbox{ if }k\leq n\\\
0&\mbox{ otherwise}
\end{cases}$. So $T$ is compact.   
For the adjoint, write $T^*=S^*\circ M^*=S^*\circ M$, and look for example at the value at $(1,0,0,\ldots)$.
