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
$$S_k(I-A)=I-A^{k+1}$$
and similarly
$$(I-A)S_k=I-A^{k+1}$$
where $S_k$ is the sum of the first $k$ terms in the series.  Then it shows that
$$|A^{k+1}|\leq|A|^{k+1}$$
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, $|A|<1$?

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 $\lim_{k\to\infty}A^k\to0$.
