Does for $T \in B(X)$ with $\|T\|>1$ exist $T^{-1}$? Is it true if $\|T\|>1$, where $T \in B(X)$ for some Banach space $X$, then $T^{-1}$ exists? 
I suppose that for $\|T\|=1$ this isn't true? Because, if we suppose that inverse exists for such operators, we can look at some operator $A$ which doesn't have inverse and then neither $\frac{A}{\|A\|}$ has inverse, contradiction.
I know that if $\|T\|<1$ then $(I-T)^{-1}$ exists and it is equal to $\sum_{n=0}^{\infty} T^n$, but I don't think we can use that fact.

Invertible and nonivertible operators may have arbitrary norms: just note that $\|cT\|=|c|\|T\|$ and $T$ is ivertible if and only if $cT$ is invertible (provided $c \neq 0$)
