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

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

Mathematics
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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.
\[\sum_{0}^{infinity}(\sqrt(k+1) -\sqrt(k))/k\]
wait does it converge to 0 if it is so....i can give you an explanation

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Other answers:

The starting index should probably be 1.
Oh yeah, starting index is 1 my bad
yeah,, parth is right
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?
\[\large \sum_{k=1}^{\infty} \left(\frac{\sqrt{k+1} -\sqrt{k}}{k}\right)\]
\[\sqrt{k-1/k} -1\] \[\sqrt{1-1/k} -1\] if k goes to infinity 1/k goes to zero then \[\sqrt{1}-1=0\]
but you've got to show the convergence of the series
Yeah, it's not enough to show the sequence goes to 0, otherwise 1/n would converge.
Unless I'm missing something about what you're saying.
Could you perhaps use a ratio test to show convergence?
i think i should that on ... do i have to use the summation sign?
oh wait... i got what ur saying now
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
\[\lim_{k\rightarrow \infty} \sqrt[n]{|a_n|} \ne 1\] The root test might work too.
whoop, meant o write n instead of k*
Oh right didn't think of the root test. That might work out more nicely.
So you'd end up with some k^(k+1/2) type stuff, all of which would go to 1 right?
|dw:1443515996560:dw|Ugly drawing, but because the tops both go to 1, it'll go to 0/1, meaning it converges?
Is that right?

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