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2015 Kazakhstan National Olympiad
1
Bounding the sum of squares of reciprocals
Bounding the sum of squares of reciprocals
Source:
January 9, 2015
inequalities
inequalities proposed
Kazakhstan
Problem Statement
Prove that
1
2
2
+
1
3
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⋯
+
1
(
n
+
1
)
2
<
n
⋅
(
1
−
1
2
n
)
.
\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{(n+1)^2} < n \cdot \left(1-\frac{1}{\sqrt[n]{2}}\right).
2
2
1
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1
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⋯
+
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n
+
1
)
2
1
<
n
⋅
(
1
−
n
2
1
)
.
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