Prove: √(2n) / ∑( 1/√(2n+1) - (1/2n) ) converges to 1 as n -> ∞?
This question is inspired by another one posted by KT:
Of course, the sum is from k = 1 to k = n, and the question is more accurately stated as: Prove that √(2n) / ∑(k = 1 to n) ( 1/√(2k+1) - (1/2k) ) converges to 1 as n -> ∞
Zeta, you are correct, the absolute difference increases without bound, which is why I didn't ask for proof that √(2n) is an asymptote. Nonetheless, the ratio does converge to 1.
That was pretty neat, Zo Maar
First Grade Rocks!, a slightly different way is to show that as the n+1 term is added, increasing the sum by 1/√(2n+3) - 1/(2n+2), it approaches the difference √(2n+2) - √(2n) for large n.
- Zo MaarLv 51 decade agoFavourite answer
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.
- 1 decade ago
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
2-1 = 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.
- 1 decade ago
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!).