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3
Integral part of a sum
Integral part of a sum
Source: Romanian TST 2018 Problem 3 Day 4
May 25, 2020
floor function
algebra
inequalities
Problem Statement
Given an integer
n
≥
2
n \geq 2
n
≥
2
determine the integral part of the number
∑
k
=
1
n
−
1
1
(
1
+
1
n
)
…
(
1
+
k
n
)
\sum_{k=1}^{n-1} \frac {1} {({1+\frac{1} {n}}) \dots ({1+\frac {k} {n})}}
∑
k
=
1
n
−
1
(
1
+
n
1
)
…
(
1
+
n
k
)
1
-
∑
k
=
1
n
−
1
(
1
−
1
n
)
…
(
1
−
k
n
)
\sum_{k=1}^{n-1} (1-\frac {1} {n}) \dots(1-\frac{k}{n})
∑
k
=
1
n
−
1
(
1
−
n
1
)
…
(
1
−
n
k
)
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