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Rafial
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I've been asked to answer this question but I'm not sure what way to go about it. It's a proof question. I'd appreciate any help, thanks
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Smaug123
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(Original post by Rafial)
I've been asked to answer this question but I'm not sure what way to go about it. It's a proof question. I'd appreciate any help, thanks
I immediately know the answer - the second thing I tried worked. If it's false, you are looking for a sequence which only just doesn't converge - what's the most obvious example of that, and does it work as a counterexample? If it's true, you're going to have to use that the sum of a_n^2 and b_n^2 converge, but there's no obvious way to link that to this sum; however, by absolute convergence, you can split up and reorder sums. Does that help?
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poorform
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This was the first thing that came to mind when I saw this problem.

I know it's false but can someone explain why?

we know the sums (a_n)^2 and (b_n)^2 converge to some real numbers x and y say. Since they are a sequence of non negative terms then we can say the sum (a_n)^2 + (b_n)^2 converges to x+y by application of the algebra of limits to the partial sums of the respective series. (not sure if this is valid)

But it is obvious that the sum of ((a_n) + (b_n))/n^(1/3) < (a_n)^2 and (b_n)^2 for all natural numbers.

Hence by the comparison test we have that ((a_n) + (b_n))/n^(1/3)) converges.

thanks
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DFranklin
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(Original post by poorform)
This was the first thing that came to mind when I saw this problem.

I know it's false but can someone explain why?

we know the sums (a_n)^2 and (b_n)^2 converge to some real numbers x and y say. Since they are a sequence of non negative terms then we can say the sum (a_n)^2 + (b_n)^2 converges to x+y by application of the algebra of limits to the partial sums of the respective series. (not sure if this is valid)

But it is obvious that the sum of ((a_n) + (b_n))/n^(1/3) < (a_n)^2 and (b_n)^2 for all natural numbers.
What happens if a=0.1, b=0.1, n = 1?
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poorform
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Okay thanks that helps me see where the answer is flawed, then I don't really know how to go about this problem.
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