Exercise :
Let $X$ be a Banach space, $T \in B(X)$ which means that $T$ is a bounded linear operator $T : X \to X$. We suppose that for all $y \in X$ the series $$\sum_{n=1}^\infty\|T^ny\|$$ converges. Show that for all $y \in X$, the equation $x = y + Tx$ has a unique solution.
Attempt/Thoughts :
Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n \to T$ as $n\to \infty$ over $B(X)$.
Since the given series is convergent, this means that :
$$\Bigg|\sum_{n=k+1}^\infty\|T^ny\|\Bigg| < \epsilon $$
Also, since $T$ is a bounded operator, this means that :
$$\|Ty\| \leq M \|y\|$$ I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !
Formally, $x$ would be given by $$ x "=" (1-T)^{-1} y$$ where by the formula for a geometric series, we should have $$ (1-T)^{-1} "=" \sum_{n=0}^\infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := \sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=\sum_0^\infty T^n y$, and moreover $$(1-T)^{-1}y := \sum_{n=0}^\infty T^n y$$ is a bounded map in $B(X)$. Then just compute (since $T\in B(X)$) $$ (1-T) x = \lim_{N\to\infty }(1-T) x_N = \lim_{N\to\infty} y - T^{N+1}y = y$$ Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$, $$(1-T)^{-1} y = x.$$ This proves that $1-T$ is invertible, and uniqueness follows.