Show that the spectrum of an operator on $\ell^2(\mathbb{N})$ is $\{0\}$. The problem I have comes from Walter Rudin's Functional Analysis, chapter 10 exercise 19.  The exercise begins with the following:
Let $S_R$ be the right shift operator, acting on $\ell^2=\ell^2(\mathbb{N})$.  Let $\{c_n\}$ be a sequence of complex numbers such that $c_n\neq0$ but $c_n\to0$ as $n\to\infty$.
Define $M\in\mathcal{B}(\ell^2)$ by
\begin{align*}
(Mf)(n)=c_nf(n)\qquad(n\in\mathbb{N}),
\end{align*}
and define $T\in\mathcal{B}(\ell^2)$ by $T=MS_R$.
I need to show that the spectrum $\sigma(T)$ of $T$ is $\{0\}$.  So far I have shown that the operator norm of $T^m$ is $b_m=\sup\{|c_{n-m+1}\cdots c_n|:n>m\}$.  Since the spectral radius $\rho(T)$ is the limit of $\|T^m\|^{1/m}$, as $m\to\infty$, I will be done if I can show that $(b_m)^{1/m}\to0$ as $m\to\infty$.  However, I am having trouble showing this.    
Any hints as to how I should proceed would be greatly appreciated.
 A: Notice that, the way you have defined $T=MS_R$, we have
\begin{equation}T^1(a_1,a_2,a_3,\cdots)=(0,c_2a_1,c_3a_2,c_4a_3,\cdots),\end{equation}
\begin{equation}T^2(a_1,a_2,a_3,\cdots)=(0,0,c_3c_2a_1,c_4c_3a_2,c_5c_4a_3,\cdots),\end{equation}
and so on.
In particular, if $(e_i)_{i=1}^\infty$ is the canonical basis for $\ell_p$, $p=2$, then we have
\begin{equation}T^m\sum_{i=1}^\infty a_ie_i=\sum_{i=1}^\infty \left(\sum_{j=1+i}^{m+i}c_j\right)a_ie_{m+i}\end{equation}
and hence, by Holder and symmetry of $(e_i)_{i=1}^\infty$,
\begin{equation}\|T^m\sum_{i=1}^\infty a_ie_i\|=\|\sum_{i=1}^\infty \left(\prod_{j=1+i}^{m+i}c_j\right)a_ie_{m+i}\|\leq\sup_{i\in\mathbb{N}}|\prod_{j=1+i}^{m+i}c_j|\|\sum_{i=1}^\infty a_ie_i\|\end{equation}
Thus,
\begin{equation}\|T^m\|\leq\sup_{i\in\mathbb{N}}\prod_{j=1+i}^{m+i}|c_j|\end{equation}
Now pick any $\epsilon>0$, find $N\in\mathbb{N}$ so that $|c_n|<\epsilon$ for all $n\geq N$, and write
\begin{equation}K=\sup_{i\in\mathbb{N}}\prod_{j=1+i}^{N+i}|c_j|.\end{equation}
Then
\begin{equation}\|T^{N+m}\|\leq\sup_{i\in\mathbb{N}}\prod_{j=1+i}^{N+m+i}|c_j|\leq\sup_{i\in\mathbb{N}}\prod_{j=1+i}^{N+i}|c_j|\prod_{j=N+i+1}^{N+i+m}|c_j|\leq \epsilon^m K=\epsilon^{N+m}(\epsilon^{-N}K).\end{equation}
Thus, $\|T^{N+m}\|^{1/(N+m)}\leq (\epsilon^{-N}K)^{1/(N+m)}\epsilon\to\epsilon$ as $m\to\infty$.  Since $\epsilon>0$ was arbitrary, we have what we want.
