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

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

Rosa Maria Gtz.
  • 3,756

2 Answers2

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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.
Community
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user1551
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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?).
Clement C.
  • 67,323
  • 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}

     – Rosa Maria Gtz. 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} – Rosa Maria Gtz. Jun 27 '13 at 22:35
  • now i do not how conclude that converge or diverge? – Rosa Maria Gtz. 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
  • Ok you said that i need to compute the norm of \begin{bmatrix}(-1)^n / n^2 & (-1)^{(n+1)}/n \ 0 & (-1)^n / n^2 \end{bmatrix} then the $$ \sum_{n=0}^{\infty} $$ \begin{bmatrix}(-1)^n / n^2 & (-1)^{(n+1)}/n \ 0 & (-1)^n / n^2 \end{bmatrix} – Rosa Maria Gtz. Jun 27 '13 at 22:52
  • will converge or diverge depending the norm? – Rosa Maria Gtz. Jun 27 '13 at 22:53
  • i can not conclude that... – Rosa Maria Gtz. Jun 27 '13 at 23:03
  • All I'm saying is that if the norm of $A^n/n^2$ ends up defining a converging series, then $A^n/n^2$ is a series which converges normally. That's the definition. And if each of the four coefficients of $A^n/n^2$ defines a converging series, then $A^n/n^2$ itself is a converging series (at least for the weakest type of convergence). – Clement C. Jun 28 '13 at 02:53
  • yea you are right but i can prove only compute the series of each aij but thanks! – Rosa Maria Gtz. Jun 28 '13 at 04:22
  • Take any norm you want (the sup of entries, for instance): it converges for one norm iff it converges for all norms. This is true because it's a space with finite dimension. – Clement C. Jun 28 '13 at 18:30