Adamm:
Since the question is in English I might as well answer it in English.
Also idk who wrote that but they should learn some proper LaTeX probably.
| | 1 | |
For simplicity let ai = |
| . We have |
| | √4n2−(i−1)α | |
| | (x+a1)...(x+an)−xn | |
((x+a1)...(x+an))1/n−x = |
| |
| | ∑ ((x+a1)...(x+an))i/nxn−1−i | |
where i = 0, ..., n−1.
Dividing both nominator and denominator by x
n−1, clearly the nominator will converge to
a
1+...+a
n since (x+a
1)...(x+a
n)−x
n = (a
1+...+a
n)x
n−1+w(x) where w(x) is a polynomial
of degree < n−1.
On the other hand the denominator becomes ∑ ((1+a
1/x)...(1+a
n/x))
i/n → n.
| | 1 | |
Hence n*Sn = ∑ |
| . |
| | √4n2−(i−1)α | |
| | 1 | | n | | 1 | |
We have |
| ≤ n*Sn ≤ |
| → |
| for α < 2. |
| | 2 | | √4n2−(n−1)α | | 2 | |
| | 1 | | 1 | | 1 | |
For α = 2 we have n*Sn = ∑ |
| = |
| ∑ |
| |
| | √4n2−(i−1)2 | | n | | √4−(i−1)2/n2 | |
| | 1 | |
and so clearly n*Sn → ∫01 |
| dx = arcsin(1/2)−arcsin(0) = π/2. |
| | √4−x2 | |
Then [1/2+π/2] = 2.
