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# Rigorous Convergence Proof watch

1. How would I prove rigorously that

e.g.
an = n-x converges to 0 where x > 0 as n -> ∞. Prove rigorously using "epsilon" and N.

I've written, Given "epsilon" >0, aN <= "epsilon"^x if N >= 1 / "epsilon"^x.

Choosing N = 1 / "epsilon"^x

Then \forall nx > N, | an | < "epsilon"^x

I don't think I've done correctly, I'm not sure, could anyone help please? Also an explaining this would be a great help!

Thanks
2. Okay, you want

N-x < e

Raise this to the power -1/x, remembering to switch the inequality around because you've raised to a negative power.

N > e-1/x = (1/e)1/x

So if you choose n>N, then you'll certainly have n-x<e.

Your chosen N doesn't work for x<1, I think.
3. Thank you!

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Updated: November 22, 2005
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