convergence of a series in the space of bounded linear operator I need help in showing that: 
If $X$ is a Banach space and $T \in L(X,X)$ have $||T||<1$. Use the completeness of $L(X,X)$ to show that $\sum_{n=0}^{\infty}T^n$ converges in $L(X,X)$. where $L(X,X)$
is the space of bounded linear operators. 
thanx in advance.   
 A: For any non-negative integer $m$ let
$S_m = \sum_{n = 0}^m T^n; \tag{1}$
we will show that the sequence $S_m$ is Cauchy.  Let $k > l$ be non-negative integers.  Then
$S_k - S_l = \sum_{n = l + 1}^k T^n = T^{l + 1} \sum_{n = 0}^{k -  l - 1} T^n = T^{l + 1}S_{k - l -1}; \tag{2}$
thus
$\Vert S_k - S_l \Vert = \Vert T^{l + 1}S_{k - l -1} \Vert \le \Vert T^{l + 1} \Vert \Vert S_{k - l -1} \Vert \le \Vert T \Vert^{l + 1} \Vert S_{k - l -1} \Vert. \tag{3}$
Consider $\Vert S_{k - l -1} \Vert$; we have
$\Vert S_{k - l -1} \Vert = \Vert \sum_{n = 0}^{k -  l - 1} T^n \Vert \le \sum_{n = 0}^{k -  l - 1} \Vert T^n \Vert \le \sum_{n = 0}^{k -  l - 1} \Vert T \Vert^n$
$\le \sum_{n = 0}^\infty \Vert T \Vert^n = \dfrac{1}{1 - \Vert T \Vert}, \tag{4}$
where we know everything converges since $\sum_{n = 0}^\infty \Vert T \Vert^n$ is a real geometric series with ratio $\Vert T \Vert < 1$; it is well known that the sum of such a series is $(1 - \Vert T \Vert)^{-1}$.  Using (4) in (3) we find
$\Vert S_k - S_l \Vert \le \dfrac{\Vert T \Vert^{l + 1}}{(1 - \Vert T \Vert)}; \tag{5}$
again recalling that $\Vert T \Vert < 1$, we see that it follows from (5) that $\Vert S_k - S_l \Vert$ may be made arbitrarily small by taking $l$ sufficiently large, independently of $k$.  Thus $S_m$ is a Cauchy sequence in $L(X, X)$; as such, since $L(X, X)$ is complete, the limit $S = \sum_{n = 0}^\infty T^n \in L(X, X)$; the infinite series converges to $S$ in $L(X, X)$.  QED.
Nota Bene:  It is worth observing, as is hinted in the above, that
$(I - T)^{-1} = \sum_{n = 0}^\infty T^n; \tag{6}$
this may be seen as follows:
$(I - T)\sum_{n = 0}^m T^n = I - T^{m + 1} \tag{7}$
by simple algebra; now if we let $m \to \infty$ the right-hand side approaches $I$, since $T^{m + 1} \to 0$ by virtue of $\Vert T \Vert < 1$ ($\Vert T^{m + 1} \Vert \le \Vert T \Vert^{m + 1}$).  This is a well-known and exceedingly useful identity in the theory of linear operators.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
