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Showing $sum_k geq 0 kA^k$ converges while $vert vert Avert vert
The Next CEO of Stack OverflowIf $sum_n geq 1X_n$ converges a.s. then $forall a > 0: sum P(|X_n|>a) < infty$Are there norms on $BbbC^m$ and $BbbC^n$ so that the norm $VertcdotVert$ is a subordinate norm?If $ sum_n=1^inftyx_na_n $ converges when $x_nto 0,$ then $ sum_n=1^inftya_n $ also converges.Prove that the series $displaystyle sum_n=1^infty a_n$ converges in $X$.Theorem 3.55 in Baby Rudin: Every re-arrangement of an absolutely convergent series converges to the same sum in every normed space?Theorem 3.22 in Baby Rudin: Is this proof correct?Hints on showing Cauchy sequence convergesIf $(2x_n+1-x_n)$ converges to $x$, then show that $(x_n)$ converges to $x$.If $leftVert ArightVert geq c$ then $left|lambdaright|>c$ for all eigenvalues of $A$Show directly that if $s_n$ is a Cauchy sequence then so is $s_n$. Conclude that $s_n$ converges whenever $s_n$ converges.
$begingroup$
Let $vert vert cdot vert vert$ be a matrix norm on $A$ where $vert vert Avert vert < 1$. Show that $sum_k geq 0 k A^k$ converges.
My ideas: Let $m<l$
$1.$ Let $vertvertsum_k=0^lkA^k-sum_k=0^mkA^kvertvert=vertvertsum_k=m+1^lkA^kvertvertleq sum_k=m+1^lvertvert kA^kvertvert=sum_k=m+1^lvert kvert vert vert A^kvertvertleq sum_k=m+1^lvert kvert vert vert Avertvert^k$
If I can remove $vert k vert$ then I am can show that it is a cauchy sequence and subsequently a convergent sequence.
other ideas: Am I allowed to simply take the derivative of $sum_k geq 0 k A^k$, but how would I then be able to compare $sum_k geq 0 k A^k$ and $sum_k geq 0 A^k$? Looking for tips.
matrices convergence optimization
$endgroup$
add a comment |
$begingroup$
Let $vert vert cdot vert vert$ be a matrix norm on $A$ where $vert vert Avert vert < 1$. Show that $sum_k geq 0 k A^k$ converges.
My ideas: Let $m<l$
$1.$ Let $vertvertsum_k=0^lkA^k-sum_k=0^mkA^kvertvert=vertvertsum_k=m+1^lkA^kvertvertleq sum_k=m+1^lvertvert kA^kvertvert=sum_k=m+1^lvert kvert vert vert A^kvertvertleq sum_k=m+1^lvert kvert vert vert Avertvert^k$
If I can remove $vert k vert$ then I am can show that it is a cauchy sequence and subsequently a convergent sequence.
other ideas: Am I allowed to simply take the derivative of $sum_k geq 0 k A^k$, but how would I then be able to compare $sum_k geq 0 k A^k$ and $sum_k geq 0 A^k$? Looking for tips.
matrices convergence optimization
$endgroup$
add a comment |
$begingroup$
Let $vert vert cdot vert vert$ be a matrix norm on $A$ where $vert vert Avert vert < 1$. Show that $sum_k geq 0 k A^k$ converges.
My ideas: Let $m<l$
$1.$ Let $vertvertsum_k=0^lkA^k-sum_k=0^mkA^kvertvert=vertvertsum_k=m+1^lkA^kvertvertleq sum_k=m+1^lvertvert kA^kvertvert=sum_k=m+1^lvert kvert vert vert A^kvertvertleq sum_k=m+1^lvert kvert vert vert Avertvert^k$
If I can remove $vert k vert$ then I am can show that it is a cauchy sequence and subsequently a convergent sequence.
other ideas: Am I allowed to simply take the derivative of $sum_k geq 0 k A^k$, but how would I then be able to compare $sum_k geq 0 k A^k$ and $sum_k geq 0 A^k$? Looking for tips.
matrices convergence optimization
$endgroup$
Let $vert vert cdot vert vert$ be a matrix norm on $A$ where $vert vert Avert vert < 1$. Show that $sum_k geq 0 k A^k$ converges.
My ideas: Let $m<l$
$1.$ Let $vertvertsum_k=0^lkA^k-sum_k=0^mkA^kvertvert=vertvertsum_k=m+1^lkA^kvertvertleq sum_k=m+1^lvertvert kA^kvertvert=sum_k=m+1^lvert kvert vert vert A^kvertvertleq sum_k=m+1^lvert kvert vert vert Avertvert^k$
If I can remove $vert k vert$ then I am can show that it is a cauchy sequence and subsequently a convergent sequence.
other ideas: Am I allowed to simply take the derivative of $sum_k geq 0 k A^k$, but how would I then be able to compare $sum_k geq 0 k A^k$ and $sum_k geq 0 A^k$? Looking for tips.
matrices convergence optimization
matrices convergence optimization
asked Mar 18 at 19:53
SABOYSABOY
710311
710311
add a comment |
add a comment |
1 Answer
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$begingroup$
You could do it like this:
$sum_k=0^infty kcdot q^k$ converges for every $q$ with $qin(-1,1)$ (Ratio test)
$sum_k=0^infty|kA^k|leqsum_k=0^infty kcdot |A|^k<infty$
If $(x_k)$ is a sequence in a Banach space with $sum_k=0^infty|x_k|<infty$ then $sum_k=0^infty x_k$ converges
$endgroup$
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You could do it like this:
$sum_k=0^infty kcdot q^k$ converges for every $q$ with $qin(-1,1)$ (Ratio test)
$sum_k=0^infty|kA^k|leqsum_k=0^infty kcdot |A|^k<infty$
If $(x_k)$ is a sequence in a Banach space with $sum_k=0^infty|x_k|<infty$ then $sum_k=0^infty x_k$ converges
$endgroup$
add a comment |
$begingroup$
You could do it like this:
$sum_k=0^infty kcdot q^k$ converges for every $q$ with $qin(-1,1)$ (Ratio test)
$sum_k=0^infty|kA^k|leqsum_k=0^infty kcdot |A|^k<infty$
If $(x_k)$ is a sequence in a Banach space with $sum_k=0^infty|x_k|<infty$ then $sum_k=0^infty x_k$ converges
$endgroup$
add a comment |
$begingroup$
You could do it like this:
$sum_k=0^infty kcdot q^k$ converges for every $q$ with $qin(-1,1)$ (Ratio test)
$sum_k=0^infty|kA^k|leqsum_k=0^infty kcdot |A|^k<infty$
If $(x_k)$ is a sequence in a Banach space with $sum_k=0^infty|x_k|<infty$ then $sum_k=0^infty x_k$ converges
$endgroup$
You could do it like this:
$sum_k=0^infty kcdot q^k$ converges for every $q$ with $qin(-1,1)$ (Ratio test)
$sum_k=0^infty|kA^k|leqsum_k=0^infty kcdot |A|^k<infty$
If $(x_k)$ is a sequence in a Banach space with $sum_k=0^infty|x_k|<infty$ then $sum_k=0^infty x_k$ converges
edited 2 days ago
answered Mar 18 at 20:08
triitrii
80817
80817
add a comment |
add a comment |
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