# Check the convergence of the series of matrices

$$\sum_{k=1}^{\infty} ( 1/ k^2 ) A ^k$$ where

A =\begin{bmatrix}-1 & 1 \\ 0 & -1 \end{bmatrix}

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Your first equation is weird, did you mean $\sum_{k=0}^{\infty}\frac{1}{k^2}A^k$? Also, please let us know what have you tried and where you are stuck at. – Lord Soth Jun 27 '13 at 22:02
ok iñm writting all i did – Knight Jun 27 '13 at 22:06
$k$ should run from $1$ to infinity, not $0$ to infinity. – user1551 Jun 27 '13 at 22:17
I do not how i can conclude that converge or diverge? – Knight Jun 27 '13 at 22:23

Hints.

1. Let $J=\pmatrix{0&1\\ 0&0}$. Then $A=-I+J$. Since $J^2=0$, in the binomial expansion of $A^k = (-I+J)^k$, only two terms remain.
2. The value of $\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$ is known.
3. The value of $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}$ is also known.
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Provided Lord Soth's interpretation of your question (cf. comment above) is correct, here's an approach which'll help you get what you want:

• Show (by induction, or any means) that $A^k$ has a "nice" expression, of the form $$A^k = (-1)^k\begin{pmatrix} 1 & \varphi(k) \\ 0 & 1 \end{pmatrix}$$ for some very convenient function $\varphi$;
• Show that $\frac{(-1)^k\varphi(k)}{k^2}$ is a convergent or divergent series. Conclude about the series $\frac{A^k}{k^2}$ (convergence? If so, type of convergence?).
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I prove that A^n = \begin{bmatrix}-1 & 1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix}(-1)^n & n(-1)^(n+1) \\ 0 & (-1)^n \end{bmatrix} and then $$\sum_{n=0}^{\infty} 1/ n^2 = ( 1/ n^2 ) A ^n$$ = $$\sum_{n=0}^{\infty}$$ \begin{bmatrix}(-1)^n / n^2 & (-1)^(n+1)/n \\ 0 & (-1)^n / n^2 \end{bmatrix} – Knight Jun 27 '13 at 22:24
I can't read or understand what you typed... I think your expression for $A^n$ is correct, but I cannot be sure unless you correct/fix the $\LaTeX$ code. – Clement C. Jun 27 '13 at 22:29
I prove that $$\sum_{n=0}^{\infty} ( 1/ n^2 ) A ^n =$$ \begin{bmatrix}(-1)^n / n^2 & (-1)^{(n+1)}/n \\ 0 & (-1)^n / n^2 \end{bmatrix} – Knight Jun 27 '13 at 22:35
now i do not how conclude that converge or diverge? – Knight Jun 27 '13 at 22:37
OK. Now, you can conclude (the matrix series converges if it converges coefficient-wise; it converges normally if the series of the norms converges). (Check what it means for a series of matrix to converge.) – Clement C. Jun 27 '13 at 22:37