A criterion for quasi-nilpotent operators to be nilpotent Let $n$ be a fixed positive integer, $X$ a Banach space and $M : X \to X$ a bounded linear operator, then apparently $M$ being nilpotent, that is $M^n = 0$, is the same as it being quasi-nilpotent $\lim_{k \to \infty} \|M^k\|^\frac{1}{k} = 0$ and having a bound on the "growth" of the resolvent, to be precise that $|\lambda|^{n}\|(\lambda I  - M)^{-1}\| <\infty$, for all $|\lambda| < 1, \lambda \neq 0.$
I have managed to prove one direction, which is that $T^n = 0$ implies that second condition, clearly $\lim_{k\to \infty} \|T^k\|^\frac{1}{k} = 0$ and using the series expansion for the resolvent $\lambda^n (\lambda I  - M)^{-1} = \lambda^n \sum_{k=0}^{n-1} \frac{M^k}{\lambda^{k+1}}$, from which it follows:
$$|\lambda|^{n}\|(\lambda I  - M)^{-1}\|  = \left\|  \lambda^n \sum_{k=0}^{n-1} \frac{M^k}{\lambda^{k+1}}\right\| \leq   \sum_{k=0}^{n-1} \left\| 
 \lambda^{n-k-1} M^k \right\| \leq \sum_{k=0}^{n-1} \left\| 
 M \right\|^k < \infty $$
Which proves one direction. However, I have been unable to show the converse, how would one go about doing that?
 A: First, since $\|M^k\|^{1/k} \rightarrow 0$, for all $\lambda$, the series $\sum_k{\lambda^kM^k}$ is normally convergent, let’s call $T_{\lambda}$ this operator, which is a holomorphic function of $\lambda$. Moreover, we know that $T_{\lambda}=\lambda^{-1}R(\lambda^{-1})$ (where $R(\lambda)=(\lambda I-M)^{-1}$), so that (by the second assumption) $\|T_{\lambda}\| = O(\lambda^{n-1})$ (as $\lambda \rightarrow \infty$).
In particular, for any $x \in X$ and $u \in X^{\star}$, $\lambda \longmapsto u(T_{\lambda}x)$ is an entire function with polynomial growth of degree at most $n-1$, hence is a polynomial of degree at most $n-1$, and therefore (because its Taylor series at zero is clear), we see that $uM^n(x)=0$ for all $x,u$, which implies that $M^{n}=0$.
Edit: we don’t even actually need to use $x,u$. Because of the series, we have (wrt the operator topology), for $r>0$, $M^n=r^{-n}\int_0^{1}{e^{-2i\pi n t}T_{re^{2i\pi t}}\,dt}$, which implies (by the norm bound as $r \rightarrow \infty$) that $M^n=0$.
