Geometric series of matrices

I am currently reading ‘Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach’ by J. Hubbard and B. Hubbard. In the first chapter, there is the proposition:

Let A be a square matrix. If $|A|<1$, the series $$S=I+A+A^2+\cdots$$
converges to $(I-A)^{-1}$.

The proof first shows that
and similarly
where $S_k$ is the sum of the first $k$ terms in the series. Then it shows that
and according to the proof in the book, it can be said from this that $\lim_{k\to \infty}A^{k+1}=0$ when $|A|<1$. Consequently, $S(I-A)=I$ and $(I-A)S=I$. Therefore $S=(I-A)^{-1}$.

However I do not understand how it can be said that $\lim_{k\to \infty}A^{k+1}=0$ when $|A|<1$. In addition, all similar propositions I have found on the internet state $|\lambda_i|<1$ as the necessary and sufficient condition. How does this relate to the condition, $|A|<1$?

Solutions Collecting From Web of "Geometric series of matrices"

Since $|A|<1$, and since you state that $\left|A^k\right|\le|A|^k$, you clearly have $\lim_{k\to\infty}\left|A^k\right|\to0$.

I don’t know which norm you are using, but for every norm this also means $\lim_{k\to\infty}A^k\to0$.

It is an answer to the last question: how does the condition (P): for every $i$, $|\lambda_i|<1$ relate to the condition, $||A||<1$ ?

Firstly $||A||<1$ implies (P) and, in general, the converse is false.

More precisely, let $\rho(A)=\sup_i(|\lambda_i|)$ be the spectral radius of $A$. It is not true that $\rho()$ is a norm ; yet it is almost true. Indeed, for every $\epsilon$, there is a matricial norm s.t. $||A||<\rho(A)+\epsilon$ (note that the proof of the previous result is not obvious). Thus, if $\rho(A)<1$, then there is a matricial norm s.t. $||A||<1$.

Note that $\sum_{k=0}^\infty A^k$ may converge even if $||A||>1$ for a given norm.

Note that

if a series $\sum_{k=0}^{\infty}a_n$ converges then $\lim_{n\to \infty} a_n =0$.