Question

Series, how do I show (sqrt(k+1)-sqrt(k))/k converges?

Answer 1

I imagine this is really simple, probably a comparison test, all the other problems in this section have been 1 or 2 steps, I'm just not seeing it.

Answer 2

\[\sum_{0}^{infinity}(\sqrt(k+1) -\sqrt(k))/k\]

Answer 3

wait does it converge to 0 if it is so....i can give you an explanation

Answer 4

The starting index should probably be 1.

Answer 5

Oh yeah, starting index is 1 my bad

Answer 6

yeah,, parth is right

Answer 7

And uhh, the sequence goes to 0. If it didn't then the problem would be easy, would just say "lim =/= 0 therefore diverges"

But the series doesn't converge to 0 I don't think?

Answer 8

\[\large \sum_{k=1}^{\infty} \left(\frac{\sqrt{k+1} -\sqrt{k}}{k}\right)\]

Answer 9

\[\sqrt{k-1/k} -1\]
\[\sqrt{1-1/k} -1\]
if k goes to infinity
1/k goes to zero
then
\[\sqrt{1}-1=0\]

Answer 10

but you've got to show the convergence of the series

Answer 11

Yeah, it's not enough to show the sequence goes to 0, otherwise 1/n would converge.

Answer 12

Unless I'm missing something about what you're saying.

Answer 13

Could you perhaps use a ratio test to show convergence?

Answer 14

i think i should that on ... do i have to use the summation sign?

Answer 15

oh wait... i got what ur saying now

Answer 16

Ratio test would end up with a bunch of n+1's around right? None of which would cancel nicely? Maybe you're right, I'll jot some notes real quick and see if anything is obvious

Answer 17

\[\lim_{k\rightarrow \infty} \sqrt[n]{|a_n|} \ne 1\] The root test might work too.

Answer 18

whoop, meant o write n instead of k*

Answer 19

Oh right didn't think of the root test. That might work out more nicely.

Answer 20

So you'd end up with some k^(k+1/2) type stuff, all of which would go to 1 right?

Answer 21

Ugly drawing, but because the tops both go to 1, it'll go to 0/1, meaning it converges?

Answer 22

Is that right?