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Putnam
1969 Putnam
A4
Putnam 1969 A4
Putnam 1969 A4
Source: Putnam 1969
March 31, 2022
Putnam
integration
Summation
Problem Statement
Show that
∫
0
1
x
x
d
x
=
∑
n
=
1
∞
(
−
1
)
n
+
1
n
n
.
\int_{0}^{1} x^{x} \, dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^n }.
∫
0
1
x
x
d
x
=
n
=
1
∑
∞
n
n
(
−
1
)
n
+
1
.
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