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Tournament Of Towns
1990 Tournament Of Towns
(267) 1
TOT 267 1990 Autumn A J1 compare nested fractions
TOT 267 1990 Autumn A J1 compare nested fractions
Source:
June 8, 2024
algebra
inequalities
Problem Statement
Given
a
=
1
2
+
1
3
+
1
.
.
.
+
.
.
.
99
,
a
n
d
b
=
1
2
+
1
3
+
1
.
.
.
+
.
.
.
99
+
1
100
a=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{...+\dfrac{...}{99}}}}, \,\,and\,\,\, b=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{...+\dfrac{...}{99+\dfrac{1}{100}}}}}
a
=
2
+
3
+
...
+
99
...
1
1
1
,
an
d
b
=
2
+
3
+
...
+
99
+
100
1
...
1
1
1
Prove that
∣
a
−
b
∣
<
1
99
!
100
!
|a-b| <\frac{1}{99! 100!}
∣
a
−
b
∣
<
99
!
100
!
1
(G Galperin, Moscow)
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