## anonymous one year ago Series, how do I show (sqrt(k+1)-sqrt(k))/k converges?

1. anonymous

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.

2. anonymous

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

3. anonymous

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

4. ParthKohli

The starting index should probably be 1.

5. anonymous

Oh yeah, starting index is 1 my bad

6. anonymous

yeah,, parth is right

7. anonymous

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?

8. anonymous

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

9. anonymous

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

10. ParthKohli

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

11. anonymous

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

12. anonymous

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

13. anonymous

Could you perhaps use a ratio test to show convergence?

14. anonymous

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

15. anonymous

oh wait... i got what ur saying now

16. anonymous

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

17. anonymous

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

18. anonymous

whoop, meant o write n instead of k*

19. anonymous

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

20. anonymous

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

21. anonymous

|dw:1443515996560:dw|Ugly drawing, but because the tops both go to 1, it'll go to 0/1, meaning it converges?

22. anonymous

Is that right?