Let $X$ be a real Banach space and $\, T:X \rightarrow X$ a linear bounded operator such that
$$ \Vert T \Vert_{\mathcal{B}(X)} < 1 $$
(i) $\quad$ I am asked to find the limit of $P_n = (I+T)(I+T^2)(I+T^4)\dots(I+T^{2^n})$.
I have proven that $P_n$ converges in $\mathcal{B}(X)$, but I cannot find its limit. Could you help me?
(ii) $\quad$I am asked to prove that $\forall r \in \mathbb{R}: \vert r \vert > \Vert T \Vert_{\mathcal{B}(X)} $ the operator $rI-T$ is invertible.
The only solution I can imagine is trying to apply Fredholm's Alternative. But I cannot prove the compactness of $T$. Have you any other idea?
About part (i), setting $a = \Vert T \Vert_{\mathcal{B}(X)} \in [0,1)$, I got:
$$
\Vert P_n \Vert \le \prod^n_{j=0} \big( 1 + a^{2^j} \big) = \exp \Big[ \sum^n_{j=0} \log{\big( 1 + a^{2^j} \big)} \Big] \le \exp \Big[ \sum^n_{j=0} a^j \Big]
$$
And the last quantity converges (to $M=e^{1/(1-a)}$) as $n \rightarrow \infty$.
Therefore, $\{P_n\}_{n \in \mathbb{N}}$ is bounded (by $M$) and it is Cauchy, since:
$$
\Vert P_n-P_{n-1} \Vert = \Vert T^{2^n} P_{n-1} \Vert \le \Vert T \Vert^{2^n} \Vert P_{n-1} \Vert \le a^{2^n}M \rightarrow 0
$$
As $\Big( \mathcal{B}(X), \Vert . \Vert_{\mathcal{B}(X)} \Big)$ is a Banach space, $\{P_n\}_{n \in \mathbb{N}}$ converges.
Regarding first question, you have
$$\begin{aligned} (I-T)P_n&= (I-T)(I+T)(I+T^2)(I+T^4)\dots(I+T^{2^n})\\ &= (I-T^2)(I+T^2)(I+T^4)\dots(I+T^{2^n}))\\ &= \cdots\\ &=(I-T^{2^n})(I+T^{2^n})\\ &=(I-T^{2^{n+1}}) \end{aligned}$$
As $\Vert T \Vert_{\mathcal{B}(X)} < 1$, $T^{2^{n+1}} \to 0$ and $(I-T)P_n \to I$. From there, you can conclude that $P_n \to (I-T)^{-1}$.
Regarding second question, use the equality $rI-T = r(I- (1/r) T)$ and the first question.