Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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
add comment

2 Answers

up vote 2 down vote accepted

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.
share|improve this answer
add comment

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?).
share|improve this answer
    
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
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.