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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{0}^{infinity}(\sqrt(k+1) \sqrt(k))/k\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait does it converge to 0 if it is so....i can give you an explanation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0The starting index should probably be 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah, starting index is 1 my bad

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah,, parth is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \sum_{k=1}^{\infty} \left(\frac{\sqrt{k+1} \sqrt{k}}{k}\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{k1/k} 1\] \[\sqrt{11/k} 1\] if k goes to infinity 1/k goes to zero then \[\sqrt{1}1=0\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0but you've got to show the convergence of the series

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it's not enough to show the sequence goes to 0, otherwise 1/n would converge.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless I'm missing something about what you're saying.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you perhaps use a ratio test to show convergence?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think i should that on ... do i have to use the summation sign?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait... i got what ur saying now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ratio 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{k\rightarrow \infty} \sqrt[n]{a_n} \ne 1\] The root test might work too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whoop, meant o write n instead of k*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh right didn't think of the root test. That might work out more nicely.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So you'd end up with some k^(k+1/2) type stuff, all of which would go to 1 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443515996560:dwUgly drawing, but because the tops both go to 1, it'll go to 0/1, meaning it converges?
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