In Murphy, exercise 2.6:
Let $A$ be a unital C$^*$-algebra. If $r(a) < 1$ and $b = (\sum_{n=0}^\infty a^{*n}a^n)^{1/2}$, show that $b \geq 1$ and $\lVert bab^{-1}\rVert < 1$....
where $r(a)$ denotes the spectral radius of $a$. Is the convergence of $\sum_{n=0}^\infty a^{*n}a^n$ an assumption of the problem, or does it follow from $r(a) < 1$? If it does not follow from $r(a) < 1$, is there a counter example? I can't seem to make any progress in either direction.
Since $r(a)<1$, from $$ r(a)=\lim_n\|a^n\|^{1/n}=\lim_n\|a^{*n}a^n\|^{1/2n} $$ it follows that there exist $c<1$ such that for all $n$ big enough, $$\|a^{*n}a^n\|\leq c^{2n}.$$ So the series always converges.
For a little more detail, given $\varepsilon>0$ with $c=r(a)+\varepsilon < 1$, there exists $n_0$ such that for all $n\geq n_0$ we have $$ |r(a)-\|a^{*n}a^n\|^{1/2n}|<\varepsilon.$$ In particular, $$ \|a^{*n}a^n\|^{1/2n}<r(a)+\varepsilon = c, $$ which gives $$ \|a^{*n}a^n\|\leq c^{2n}. $$