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National and Regional Contests
Romania Contests
Romania - Local Contests
Gheorghe Vranceanu
2005 Gheorghe Vranceanu
2005 Gheorghe Vranceanu
Part of
Gheorghe Vranceanu
Subcontests
(4)
1
4
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2
5
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4
3
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Another limit with e
lim
n
→
∞
(
(
1
+
1
/
n
)
−
n
∑
i
=
0
n
1
i
!
)
2
n
\lim_{n\to\infty } \left( (1+1/n)^{-n}\sum_{i=0}^n\frac{1}{i!} \right)^{2n}
lim
n
→
∞
(
(
1
+
1/
n
)
−
n
∑
i
=
0
n
i
!
1
)
2
n
A condition for a triangle to be isosceles
Let be a triangle
A
B
C
ABC
A
BC
and the points
E
,
F
,
M
,
N
E,F,M,N
E
,
F
,
M
,
N
positioned in this way:
E
,
F
E,F
E
,
F
on the segment
B
C
BC
BC
(excluding its endpoints),
M
M
M
on the segment
A
C
AC
A
C
(excluding its endpoints) and
N
N
N
on the segment
A
C
AC
A
C
(excluding its endpoints). Knowing that
B
A
E
BAE
B
A
E
is similar to
F
A
C
FAC
F
A
C
and that
B
E
=
B
M
,
F
C
=
C
N
,
A
M
=
A
N
,
BE=BM,FC=CN,AM=AN,
BE
=
BM
,
FC
=
CN
,
A
M
=
A
N
,
show that
A
B
C
ABC
A
BC
is isosceles.
Infimum of limits
Let be a sequence of real numbers
(
x
n
)
n
⩾
0
\left( x_n \right)_{n\geqslant 0}
(
x
n
)
n
⩾
0
with
x
0
≠
0
,
1
x_0\neq 0,1
x
0
=
0
,
1
and defined as
x
n
+
1
=
x
n
+
x
n
−
1
/
x
0
.
x_{n+1}=x_n+x_n^{-1/x_0} .
x
n
+
1
=
x
n
+
x
n
−
1/
x
0
.
a) Show that the sequence
(
x
n
⋅
n
−
x
0
1
+
x
0
)
n
⩾
0
\left( x_n\cdot n^{-\frac{x_0}{1+x_0}} \right)_{n\geqslant 0}
(
x
n
⋅
n
−
1
+
x
0
x
0
)
n
⩾
0
is convergent.b) Prove that
inf
x
0
≠
0
,
1
lim
n
→
∞
x
n
⋅
n
−
x
0
1
+
x
0
=
1.
\inf_{x_0\neq 0,1} \lim_{n\to\infty } x_n\cdot n^{-\frac{x_0}{1+x_0}} =1.
in
f
x
0
=
0
,
1
lim
n
→
∞
x
n
⋅
n
−
1
+
x
0
x
0
=
1.
3
4
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