# absolute convergence

• Jan 28th 2010, 12:56 PM
CarmineCortez
absolute convergence
I need to show Sum(I-T)^k from k=0 to infinity converges absolutely to T^(-1).
Where ||I-T|| < 1.

so I can show the inverse exists, and I know this is a geometric series but when I take the limit of (I-T)^k as k-> infinity -> I get 0...
• Jan 28th 2010, 06:25 PM
tonio
Quote:

Originally Posted by CarmineCortez
I need to show Sum(I-T)^k from k=0 to infinity converges absolutely to T^(-1).
Where ||I-T|| < 1.

so I can show the inverse exists, and I know this is a geometric series but when I take the limit of (I-T)^k as k-> infinity -> I get 0...

Hint: $\sum\limits_{n=0}^\infty r^n=\frac{1}{1-r}\,,\,\,for\,\,\,|r|<1\,,\,\,r\in\mathbb{C}$

Tonio
• Jan 28th 2010, 06:50 PM
CarmineCortez
Quote:

Originally Posted by tonio
Hint: $\sum\limits_{n=0}^\infty r^n=\frac{1}{1-r}\,,\,\,for\,\,\,|r|<1\,,\,\,r\in\mathbb{C}$

Tonio

But then it is 1/(matrix)...does that have a meaning
• Jan 29th 2010, 01:53 AM
HallsofIvy
Quote:

Originally Posted by CarmineCortez
But then it is 1/(matrix)...does that have a meaning

When did you tell us that this problem involved matrices? In any case, (1/T)(T)= 1 so 1 over a matrix is the inverse matrix- the multiplicative inverse.
• Jan 29th 2010, 02:58 AM
tonio
Quote:

Originally Posted by CarmineCortez
But then it is 1/(matrix)...does that have a meaning

I had a feeling that T must be a matrix but you never pointed out this; anyway, if you're asking this question then you know there's a definition and a meaning to "convergence of matrices, their limits and etc."

In this case, it turns out that $T\cdot\sum\limits_{k=0}^\infty (1-T)^k=I$

Tonio