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Austrian-Polish
1998 Austrian-Polish Competition
4
S_m(n) \le n + m (\sqrt[4]{2^m}-1), S_m(n)=\sum \sqrt[k^2]{k^m}]
S_m(n) \le n + m (\sqrt[4]{2^m}-1), S_m(n)=\sum \sqrt[k^2]{k^m}]
Source: Austrian - Polish 1998 APMC
May 4, 2020
inequalities
Sum
algebra
Problem Statement
For positive integers
m
,
n
m, n
m
,
n
, denote
S
m
(
n
)
=
∑
1
≤
k
≤
n
[
k
m
k
2
]
S_m(n)=\sum_{1\le k \le n} \left[ \sqrt[k^2]{k^m}\right]
S
m
(
n
)
=
1
≤
k
≤
n
∑
[
k
2
k
m
]
Prove that
S
m
(
n
)
≤
n
+
m
(
2
m
4
−
1
)
S_m(n) \le n + m (\sqrt[4]{2^m}-1)
S
m
(
n
)
≤
n
+
m
(
4
2
m
−
1
)
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