linear operator from $\ell^1$ to $\ell^1$ Let $ 0 < \lambda <1 $ and define $ T: \ell^1 \to \ell^1$ with $ T(x_1,x_2,x_3, \cdots ) =(\lambda x_2 ,\lambda x_3 ,\lambda x_4 , \cdots)$.
(a) Prove that $T$ is bounded and find $||T||$.
(b) For every $ n \in \mathbb{N}$ find $||T^n||$ where $T^n=T \circ T \circ T \cdots \circ T$ where the composition of $T$ is $n-$times.
(c) Prove that the series $ \sum_{n=0}^{\infty} T^n$ converges where ($ T^0=Id$).
(d) Find the norm of the operator $S$ where $S=\sum_{n=0}^{\infty} T^n$.
I need some help for (c) and (d).
Here it is what I did in (a) and (b).
It is easy to see that $T$ is linear.
(a) For every $x=(x_n)_n \in \ell^1$ it is $ ||T(x)||_{1}^{2} =\sum_{n=2}^{\infty} \lambda |x_n| \leq \lambda \sum_{n=1}^{\infty} \lambda |x_n|= \lambda ||x||_1^{2} < + \infty$
so $T(x) \in \ell^1$ and $||T|| \leq \lambda$. Also it is $||T|| \geq \frac{||T(e_2)||_1}{||e_2||_1} = \lambda$. Thus we have that $||T||= \lambda$.
(b) Using induction we see that $ T^n(x)= (\lambda^n x_{n+1} , \lambda^n x_{n+2} , \cdots )$
So as in (a) it is$||T^n||= \lambda^n$.
Thank's in advance!
 A: Set $S=\sum T^n$. Given $x\in l_1$, formally
$$
Sx = ( \sum_{n=1}^\infty \lambda^{n-1} x_n,   \sum_{n=2}^\infty \lambda^{n-2} x_n, \sum_{n=3}^\infty \lambda^{n-3} x_n,\ldots )
$$
Since $\sum|x_i|$ is finite, $Sx$  is well defined. Moreover, $Sx$  is an element of $\ell_1$:  
$$\Vert Sx\Vert_1 

=\sum_{j=1}^\infty |  \sum_{n=j}^\infty \lambda^{n-j} x_n|
 \le
\sum_{n=1}^\infty   \sum_{j=n}^\infty \lambda^{n -1}| x_j|
  
\le \sum_{n=1}^\infty \lambda^{n-1}\Vert x\Vert_1={1\over 1-\lambda}\Vert x\Vert_1
    $$
The above also shows $\Vert S\Vert\le{1\over 1-\lambda}$.
To show that the norm of $S$ is $1\over 1-\lambda$, consider the image of the vectors
$$
w_n=(\underbrace{0,0,\cdots,0\vphantom{\textstyle{1\over n}}}_{n\text{-terms}},\underbrace{ {\textstyle {1\over n}},  {\textstyle {1\over n}},\cdots, {\textstyle {1\over n}}}_{n\text{-terms}},0,0,\cdots)
$$
We have
$$ 
\eqalign
{



T^0w_n &=\lambda^{0}  (\underbrace{ 0,0,0,\ldots,0}_{n\text{-terms}}, 
\underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots    )  \cr


T^1w_n &=\lambda^{1} ( \underbrace{  0,0, \ldots,0}_{(n-1)\text{-terms}}, \underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}},
  {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots    )  \cr

T^2w_n &=\lambda^2 ( \underbrace{ 0,\ldots,0}_{(n-2)\text{-terms}},\underbrace{   {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots    )  \cr

&\vdots\cr
T^nw_n &=\lambda^{n}  ( \underbrace{    {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots    )  ;\cr
 
}
$$
from which it follows that 
$$
\Vert Sw_n\Vert 
\ge\Bigl\Vert \sum_{j=0}^n T^jw_n\Bigr\Vert=
1+\lambda+\lambda^2+\cdots+\lambda^n={1-\lambda^{n+1}\over 1-\lambda}.
$$
Since $\Vert w_n\Vert=1$ for all $n$, and since $\lim\limits_{n\rightarrow\infty}
{1-\lambda^{n+1}\over 1-\lambda}={1\over 1-\lambda}$, we have $\Vert S\Vert\ge {1\over 1-\lambda}$. 
We have already shown that $\Vert S\Vert\le {1\over 1-\lambda}$; thus 
$\Vert S\Vert= {1\over 1-\lambda}$.
