Question

determine whether the sequence [an] diverges or converges
an = squareroot of (n^2 − 2n + 8) -n

i've tried simplifying by multiplying numerator and denominator by its conjugate but i still got divergent. it converges to some number but i don't know what

− n

Answer 1

for larger and larger values of n, this begins to resemble sqrt(n^2)-n

the question i would have is how fast is it doing that?

Answer 2

are you using a denominator of 1? its hard to see if im reading your post correctly

Answer 3

yes it's a denominator of 1.

Answer 4

it's just the (squareroot of n^2 -2n +8) -n

Answer 5

Try this: \(\dfrac{\sqrt{n^{2}-2n+8}-n}{1}\cdot\dfrac{\sqrt{n^{2}-2n+8}+n}{\sqrt{n^{2}-2n+8}+n}\).

Simplify numerator and denominator separately. Something magical may occur.

Answer 6

(-2n+8) / denominator. . I factored out a 'n' from numerator and from the squareroot. Then I reduced the fraction by canceling the n's. I took the limit of n to infinity and got #/0. :/

Answer 7

How did you manage 0 in the denominator? That's no good.

Answer 8

Oh oops. I see my mistake. I subtracted n instead of adding it

Answer 9

Thanks!

Answer 10

Did you get the non-zero limit?

Answer 11

Yes -1

Answer 12

Perfect. Tricky, wasn't it. Did you use l'Hopital?

Answer 13

No

Answer 14

That's good, since it would be a wonderous circular discouragement. It's a funny case for l'Hopital. Good work reasoning it out without that confusion.