For which values of k this expression holds true : $\sum{n=1^\infty} k^n(A^n+kA^{n+1) = kA $

Question

A= $$ \begin{bmatrix} 3 & 0 & 0 \\ 2 & 7 & 0 \\ 4 & 8 & 1\\ \end{bmatrix} $$

For which values of k this expression holds true :

$$\sum_{n=1}^\infty k^n(A^n+kA^{n+1}) = kA $$

What I did

$$\sum_{n=1}^\infty k^n(A^n+kA^{n+1}) = kA - k^{N+1}A^{N+1}$$ as $N \rightarrow \infty$ using method of differences

$-1<k<1$

But the answer is

$-1/7<k<1/7$

Any help is appreciated

score 1 · Accepted Answer · answered Sep 23 '19 at 07:15

1

The diagonal entries of $A^n$ are $1, 7^n$ and $3^n$. $k$ must be small enough to send each of those towards zero.

answered Sep 23 '19 at 07:15

Empy2

Can you give a little more explanation – AKA Death Sep 23 '19 at 07:22
What would be the diagonal entries of $k^nA^n$ ? – Empy2 Sep 23 '19 at 07:23
And if k is less than 1 and greater than -1 then it will automatically go to zero – AKA Death Sep 23 '19 at 07:24
Bit the central entry is $k^n7^n=(7k)^n$, so that won^t always go to zero. – Empy2 Sep 23 '19 at 07:26

1 Answers1