Prove boundedness of the matrix series 
Suppose $A$ is a square matrix, such that all eigenvalues of $A$ has norm strictly less than $1$, can I say $\sum_{i=k_0}^kA^{k-i}$ is bounded for all large enough $k_0$ and $k$?

From some other questions, I know using Jordan canonical form, we can write it as $X^{-1}AX=D+N$ where $D$ is a diagonal matrix and $N$ is nilpotent. Then $A^{k-i}=(X(D+N)X^{-1})^{k-i}=X(D+N)^{k-i}X^{-1}$, hence $\sum_{i=k_0}^kA^{k-i}=X\left[\sum_{i=k_0}^k(D+N)^{k-i}\right]X^{-1}$.
But I don't know how to proceed because $k-i$ can be small so I don't know how to cancel the effect of $N$. I haven't learnt the techniques about this problem before, so I appreciate if you can give a detailed explanation.
 A: More generally, let $\sum_na_nz^n$ be a complex power series with radius of convergence $R>0$ and let $A\in M_p(\mathbb{C})$ satisfying $\rho(A)<R$. Then $\sum_na_nA^n$ is convergent.
Proof: The KEY: for every $\epsilon<R-\rho(A)$, there is an induced norm $||.||$ s.t. $||A||<R-\epsilon$.
Then $\sum_n|a_n|||A^n||\leq \sum_n|a_n|||A||^n\leq \sum_n|a_n|(R-\epsilon)^n<\infty$.
In particular, for every $\epsilon>0$, there is $k_0\in\mathbb{N}$ s.t., for every $k\geq k_0$, $||\sum_{n=k_0}^ka_nA^n||<\epsilon$.
A: (1) Note that for $0<c<1$,
$$ \int_K^\infty c^xx^l =\frac{c^x x^l}{\ln\ c}\bigg|_K^\infty -\int
\frac{c^x}{\ln\ c}lx^{l-1} =\sum_{i=1}^{l+1}  (-1)^{i-1}\frac{c^x
x^{l+1-i}}{(\ln\ c)^i}\bigg|_K^\infty < \epsilon
 $$ for some $K>0$
It implies that $$ \sum_{n=K}^\infty c^n(1+n+\cdots +n^m) < \epsilon
$$
(2) Assume that $A$ has $m$-by-$m$ matrix. I will use Jordan normal form : $$ A=A_1\oplus \cdots \oplus A_s $$ where $A_i$ has size $n_i$ so that $\sum n_i=m$. And $A_i=D_i+N_i,\ D_i=d_iI,\ D_iN_i=N_iD_i$ where $$ (N_i)_{a(a+1)}=1\ {\rm and\ other\ entries\ are}\ 0$$
For calculation we assume that $s=1$ (since $A$ has finite blocks)
$$ (D+N)^n = \sum_{i=0}^n \ _nC_i D^{n-i}
N^{i} =\sum_{i=0}^{m-1} \ _nC_i D^{n-i} N^{i}$$
If $D=dI,\ c=|d|$ then $\sum_{n=K}^\infty (D+N)^n$ has entry whose
absolute value is smaller than $\epsilon$. 
