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
$\Vert S\Vert= {1\over 1-\lambda}$.