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