Question

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

Answer 1

I'm thinking this is a p series. Now what is p?

Answer 2

its from my chapter on geometric series/nth term test for divergence

Answer 3

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

Answer 4

they figure it converges to 2 + sqrt(2), how did they get there?

Answer 5

As said it converges to a/(1-r) where a=1 and r=(1/sqrt(2)) Substituting those you get 2+sqrt(2)

Answer 6

guess im having trouble with the algebra, whats the best way to approach 1- (1/sqrt2)?

Answer 7

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.

Answer 8

Or you could always start off by rationalizing the numerator and denominator by multiplying top and bottom by (1+(1/sqrt(2))

Answer 9

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!

Answer 10

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.

Answer 11

heh, still doing something wrong, i get {-2-sqrt(2)}/{-2-sqrt(2)+sqrt(2)+1} =/

Answer 12

got it finally, thanks

Answer 13

Np