Uniformly Convergent Power Series Bounded Operators

Question

I am given a bounded operator T on a Hilbert space H with $\left\vert \left\vert{T}\right\vert \right\vert < 1$. I am trying to show that \begin{align*} \sum_{n = 0}^{\infty} \left(\begin{array} 11/2\\ n\end{array}\right)T^n \end{align*} converges uniformly to $(I + T)^{1/2}$ satisfying $((I + T)^{1/2})^2 = I + T$. I am not sure how to get uniform convergence for this. I tried running through the definition of uniform convergence, but was unable to to get it. I tried looking back to undergraduate analysis and consider the function $f(x) := \sqrt{1 + x}$, but I was unable to get any inspiration from this. I would appreciate any suggestions, ideas, or anything at all.

Answer 1

$\|\sum\limits_{k=m}^{n}2^{-k}T^{k}\| \leq \sum\limits_{k=m}^{n}\|2^{-k}T^{k}\|\leq \sum\limits_{k=m}^{n}2^{-k} \to 0$ as $ n >m \to \infty$. hence the partial sums of the series $\sum\limits_{k=1}^{\infty}2^{-k}T^{k}$ form a Cauchy sequence in the space of bounded operators on $H$ with the operator norm. Since this space is complete it follows that the series converges to a bounded operator $S$. Now just verify that $(I-\frac T 2) S=S(I-\frac T 2)=I$.

Answer for the revised question: the proof is similar. The series $\sum |\binom {\frac 1 2} {n}||x|^{n}$ converges for $|x| <1$. Hence $\sum \|\binom {\frac 1 2} {n}T^{n}\| <\infty$ and we can prove,as before, that the series $\sum \binom {\frac 1 2} {n}T^{n}$ converges to a bounded operator.