Yahoo Answers: Answers and Comments for Prove: √(2n) / ∑( 1/√(2n+1)  (1/2n) ) converges to 1 as n > ∞? [Mathematics]
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From Scythian1950
enSG
Mon, 05 Oct 2009 15:51:59 +0000
3
Yahoo Answers: Answers and Comments for Prove: √(2n) / ∑( 1/√(2n+1)  (1/2n) ) converges to 1 as n > ∞? [Mathematics]
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From Zo Maar: If you do not need a rigorous prove: "for...
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Tue, 06 Oct 2009 00:50:01 +0000
If you do not need a rigorous prove: "for every ε there is such δ ...", then the answer is simple.
First terms in the sum do not play a role, so one can take k>>1 to evaluate the sum. Then
√(2k+1) = √(2k) [1+1/(4k)], 1/√(2k+1) = 1/√(2k) + O(k^(3/2)).
What plays a role is the term 1/√(2k) ,
Σ1/√(2k) > ∫ 1/√(2k) dk = √(2n) + const.
Summation of (1/2k) gives a logaritmic term.
The limit is √(2n)/[√(2n) + O(ln(n))] > 1.

From Frst Grade Rocks! Ω: I like Zo Maar solution. It is very pretty. ...
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Tue, 06 Oct 2009 14:13:58 +0000
I like Zo Maar solution. It is very pretty. But here is another less elegant way to show the equivalence.
Since we are only worried about the big guys, then if we increase n by 1, the sum should increase by the same amount.
√ (2(n+1))  √ (2(n)) = 1/√(2n +2)  1/(2(n+1))
We can ignore 1/(2(n+1)), so
√ (2(n+1))  √ (2(n)) = 1/√(2n +2)
2(n+1)  √ ((2(n))* 2(n + 1) )= 1
(2(n+1)  2√ (n^2 + n)= 1
2(n+1)  2n *√(1 + 1/n) = 1
For large n's, √(1 + 1/n) = 1 +1/2n
2n +2  2n (1+1/2n) = 1
21 = 1
We have therefore shown that if we increase n by 1, the sum will also increase by the same amount, namely √(2(n+1))  √(2n)
And since they both go to infinity and there is a one to one correspondence between them the answer is one
(a + n )/(b+n) as n > ∞ is equal to 1.

From Zeta: My observation is that the difference between ...
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Mon, 05 Oct 2009 21:59:08 +0000
My observation is that the difference between √(2n) and ∑(k = 1 to n) ( 1/√(2k+1)  (1/2k) ) increases without bound. So the ratio is not necessarily 1 in the limit. Although it's a very good approximation.
Really? I might have confused myself. I thought we have the form (a + error) / a = 1 + ∞/∞ but then then ∞/∞ might approach zero. OK I take that back! There is probably a nice way to do this with integrals.
Thumbs up to Zo Maar! He must have years of experience in this sort of problem. It made me want to learn what that big O thing means (I don't understand it!).