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Baltic Way
2000 Baltic Way
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20
Part of
2000 Baltic Way
Problems
(1)
Prove the upper and lower bound of x_n
Source: Baltic Way 2000
12/17/2010
For every positive integer
n
n
n
, let
x
n
=
(
2
n
+
1
)
(
2
n
+
3
)
⋯
(
4
n
−
1
)
(
4
n
+
1
)
(
2
n
)
(
2
n
+
2
)
⋯
(
4
n
−
2
)
(
4
n
)
x_n=\frac{(2n+1)(2n+3)\cdots (4n-1)(4n+1)}{(2n)(2n+2)\cdots (4n-2)(4n)}
x
n
=
(
2
n
)
(
2
n
+
2
)
⋯
(
4
n
−
2
)
(
4
n
)
(
2
n
+
1
)
(
2
n
+
3
)
⋯
(
4
n
−
1
)
(
4
n
+
1
)
Prove that
1
4
n
<
x
n
−
2
<
2
n
\frac{1}{4n}<x_n-\sqrt{2}<\frac{2}{n}
4
n
1
<
x
n
−
2
<
n
2
.
integration
trigonometry
logarithms
algebra proposed
algebra