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Putnam
1981 Putnam
B1
Putnam 1981 B1
Putnam 1981 B1
Source: Putnam 1981
March 31, 2022
Putnam
limit
Summation
Problem Statement
Find
lim
n
→
∞
1
n
5
∑
h
=
1
n
∑
k
=
1
n
(
5
h
4
−
18
h
2
k
2
+
5
k
4
)
.
\lim_{n\to \infty} \frac{1}{n^5 } \sum_{h=1}^{n} \sum_{k=1}^{n} (5h^4 -18h^2 k^2 +5k^4).
n
→
∞
lim
n
5
1
h
=
1
∑
n
k
=
1
∑
n
(
5
h
4
−
18
h
2
k
2
+
5
k
4
)
.
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