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anonymous

  • 5 years ago

{(1/sqrt(2))^n} from n=0 to infinity; is the series convergent or divergent? if convergent, why? and sum?

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  1. anonymous
    • 5 years ago
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    I'm thinking this is a p series. Now what is p?

  2. anonymous
    • 5 years ago
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    its from my chapter on geometric series/nth term test for divergence

  3. anonymous
    • 5 years ago
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    It's a geometric series right. Its in the form a+ar+ar^2+ar^3+ar^4+...where a=1 and r=(1/sqrt(2)) When abs(r)<1 this series converges to a/(1-r) We can use the ratio test to check: Let x=(1/sqrt(2)) Then lim n->inf [x^(n+1)]/[x^n]=x Since x<1 we know it converges

  4. anonymous
    • 5 years ago
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    they figure it converges to 2 + sqrt(2), how did they get there?

  5. anonymous
    • 5 years ago
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    As said it converges to a/(1-r) where a=1 and r=(1/sqrt(2)) Substituting those you get 2+sqrt(2)

  6. anonymous
    • 5 years ago
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    guess im having trouble with the algebra, whats the best way to approach 1- (1/sqrt2)?

  7. anonymous
    • 5 years ago
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    1/[1-(1/sqrt(2))] 1/[(sqrt(2)-1)/sqrt(2)] sqrt(2)/[sqrt(2)-1] Then rationalize the denominator by multiplying the top and bottom by -sqrt(2)-1.

  8. anonymous
    • 5 years ago
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    Or you could always start off by rationalizing the numerator and denominator by multiplying top and bottom by (1+(1/sqrt(2))

  9. anonymous
    • 5 years ago
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    i worked through your first suggestion, all makes sense up to the rationalize the denominator part, why -sqrt(2)-1? so far your a life saver though, thanks!

  10. anonymous
    • 5 years ago
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    Glad i can help. So the denominator you get is sqrt(2)-1. When you multiply by -sqrt(2)-1 you get rid of the sqrt(2). Kind like how complex conjugates work. Foil it out and see.

  11. anonymous
    • 5 years ago
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    heh, still doing something wrong, i get {-2-sqrt(2)}/{-2-sqrt(2)+sqrt(2)+1} =/

  12. anonymous
    • 5 years ago
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    got it finally, thanks

  13. anonymous
    • 5 years ago
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    Np

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