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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinking this is a p series. Now what is p?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its from my chapter on geometric series/nth term test for divergence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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/(1r) 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they figure it converges to 2 + sqrt(2), how did they get there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As said it converges to a/(1r) where a=1 and r=(1/sqrt(2)) Substituting those you get 2+sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0guess im having trouble with the algebra, whats the best way to approach 1 (1/sqrt2)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/[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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or you could always start off by rationalizing the numerator and denominator by multiplying top and bottom by (1+(1/sqrt(2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Glad 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0heh, still doing something wrong, i get {2sqrt(2)}/{2sqrt(2)+sqrt(2)+1} =/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got it finally, thanks
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