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So I have the following examples:

  1. $x_k={1 \over 4^k}; k=1,2,3,\ldots$
  2. $x_k={1 \over k+1}; k=0,1,2,\ldots$
  3. $x_k={1 \over (k+1)^2}; k=1,2,3,\ldots$

I found that these are all linearly convergent.

Can someone tell me if i did something wrong, or give me better convergence proof?

The answer should be one of these following: linear, R-linear, quadratic, or superlinear


Take a look at this definitions: enter image description here

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I think you'll find that the other three notions all imply linear convergence, so it could be that you are correct that your examples are all linearly convergent, while at the same time some of them might also be, say, quadratically convergent. – Gerry Myerson Nov 19 '12 at 2:12
Indeed, you might take it as an exercise to prove that R-linear convergence implies quadratic convergence, which implies superlinear convergence, which implies linear convergence. – Gerry Myerson Nov 19 '12 at 2:14
up vote 1 down vote accepted

For #3, $1/(k+1)^2$ divided by $1/k^2$ doesn't go to zero, so the convergence is not superlinear; it is merely linear.

For #1, compare it to the definition of R-linear convergence, and I think you'll have no trouble finding appropriate values of $M$ and $c$.

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what about #2 ? – ASROMA Nov 19 '12 at 2:34
For #2, see #3. – Gerry Myerson Nov 19 '12 at 3:10
Aren't these two sublinearly convergent since $c=1$? – Erick Wong Nov 19 '12 at 3:58
@Erick, yes. I guess I read into it that $c$ could depend on $k$, but that's not the way it's written. – Gerry Myerson Nov 19 '12 at 5:31

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