Extended discussion...
(Find a draft of the solution below!)
Query
T.A.E. nicely showed Hadamard's criterion saying that the series:
$$\sum_{k=0}^\infty A_k$$
converges for $\limsup_{k\to\infty}\|A_k\|^\frac{1}{k}<1$ and certainly diverges for $\limsup_{k\to\infty}\|A_k\|^\frac{1}{k}>1$.
Now, why does one need bounded linear functionals anyway?
Explanation
The point is basically to establish(!) convergence of the series:
$$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}\quad(|\lambda|>r_\sigma(A))$$
This is done by drifting into the realm of complex analysis via bounded linear functionals.
Now, that is why one needs bounded linear functionals anyway!
Moreover, the divergence of the series is not(!) sufficient as there might be another expression for the inverse.
As an analog problem consider the real function:
$$f(x):=\frac{1}{x^2+1}$$
That one has a power series expansion around zero with radius one - inside it converges and outside it certainly diverges. It is nevertheless analytic on the whole real line just with new power series expansions:
$$f(x)=1-x^2+x^4+\ldots\quad(-1<x<1)$$
$$f(x)=\frac12\mp\frac12(x-1)+\ldots\quad(\pm1-\sqrt{2}<x<\pm1+\sqrt{2})$$
So be careful about drawing conclusions from nonexistent a.k.a. divergent expression!!
Sketchy solution...
Problem
Denote the spectral radius and the 'Hadamard' radius by:
$$r_\sigma(A):=\sup\|\sigma(A)\|$$
$$r_H(A):=\limsup_{n\to\infty}\|A^n\|^\frac{1}{n}$$
The goal is to establish:
$$r_\sigma(A)=r_H(A)$$
Strategy
On the one hand for all $|\lambda|>r_H(A)$ the inverse is explicitely given by:
$$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}$$
by Hadamard's formula as mentioned above.
Therefore $r_H(A)\geq r_\sigma(A)$.
...
On the other hand for all $|\lambda|>r_\sigma(A)$ the inverse is analytic so as well for all bounded functionals:
$$\varphi\in\mathcal{A}':\quad\varphi\left((\lambda-A)^{-1}\right)\text{ analytic}$$
So from complex analysis the Laurent series must converge:
$$\varphi\left((\lambda-A)^{-1}\right)=\sum_{n=0}^\infty\frac{\varphi(A^n)}{\lambda^{n+1}}$$
But this means that sequence of summands is bounded for every bounded linear functional separately. By the uniform boundedness theorem this applies without them as well and so:
$$\limsup_{n\to\infty}\|A^n\|^\frac{1}{n}<|\lambda|$$
Therefore $r_\sigma(A)\geq r_H(A)$.
Especially, again by Hadamard's formula:
$$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}$$
The answer is no.
This example is due to Kakutani, found in C. E. Rickart's General Theory of Banach Algebras, page $282$.
Consider the Banach space $\ell^2$ with the canonical basis $(e_n)_n$. Define a sequence of scalars $(\alpha_n)_n$ with the relation $\alpha_{2^k(2l+1)} = e^{-k}$ for $k,l \ge 0$.
Define $A : \ell^2 \to \ell^2$ with $Ae_n = \alpha_n e_{n+1}$. We have $\|A\| = \sup_{n\in\mathbb{N}}|\alpha_n|$. Also define a sequence of operators $A_k : \ell^2 \to \ell^2$ with
$$A_k e_n = \begin{cases} 0, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\
\alpha_ne_{n+1}, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$
Then $A_k^{2^{k+1}} = 0$ so $A_k$ is nilpotent. We also have $A_k \to A$ since
$$(A - A_k)e_n = \begin{cases} e^{-k}, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\
0, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$
so $\|A - A_k\| = e^{-k} \to 0$.
For $j \in \mathbb{N}$ we have
$$A^je_n = \alpha_n\alpha_{n+1}\cdots\alpha_{n+j-1}e_{n+j}$$
Notice that $$\alpha_{1}\alpha_2\cdots\alpha_{2^t-1} = \prod_{r=1}^{t-1} \exp(-r2^{t-r-1})$$
$$r(A)= \limsup_{j\to\infty}\|A^j\|^{\frac1j} \ge \limsup_{t\to\infty} \|A^{2^t-1}\|^{\frac1{2^t-1}} \ge \limsup_{t\to\infty} \|A^{2^t-1}e_1\|_2^{\frac1{2^{t-1}}} = \limsup_{t\to\infty} |\alpha_{1}\alpha_2\cdots\alpha_{2^t-1}|^{\frac1{2^{t-1}}} \\
\ge\limsup_{t\to\infty} \left(\prod_{r=1}^{t-1} \exp(-r2^{t-r-1})\right)^{\frac1{2^{t-1}}} = \limsup_{t\to\infty}\left(\prod_{r=1}^{t-1} \exp\left(-\frac{r}{2^{r}}\right)\right) = \limsup_{t\to\infty}\exp\left(-\sum_{r=1}^{t-1}\frac{r}{2^{r}}\right) = e^{-\sum_{r=1}^\infty \frac{r}{2^{r}}}$$
Therefore $A_k \to A$ but $r(A_k) \not\to r(A)$ since $r(A_k) = 0$ but $r(A) > 0$.
Best Answer
We can apply the Banach-Steinhaus theorem (also known as the Uniform boundedness principle) to your situation as follows.
Let $X$ be a Banach space. Then we have the evaluation map from $X$ to its bidual $X''$ $$ \varphi:X\rightarrow X'', x \rightarrow \varphi(x) $$ with $$ (\varphi(x))(f) = f(x)\qquad for\ x\in X\ and\ f\in X'. $$ It is "well known" that $\varphi$ is a linear isometry (see for example here).
Now, in your situation, we are given the sequence $((\lambda x)^n)$ in $X.$ Applying $\varphi,$ we get a sequence $(\varphi((\lambda x)^n))$ in $X''.$ Suppose now we are given that $$ \forall \rho \in X'\exists M_\rho < \infty: |\rho((\lambda x)^n)| = |(\varphi((\lambda x)^n))(\rho)| \leq M_\rho, $$ i.e. the sequence $(\varphi((\lambda x)^n))$ of continuous linear operators from $X'$ to $\mathbb C$ is pointwise bounded. Then Banach-Steinhaus tells us that this sequence is norm-bounded, in symbols $$ \exists M < \infty \forall n\in \mathbb N: \|\varphi((\lambda x)^n)\|_{X''}= \|(\lambda x)^n)\|_X \leq M. $$ Here we have used that $\varphi$ is an isometry.
Thus we have shown that the original sequence $((\lambda x)^n)$ is bounded, as desired.