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Putnam
2016 Putnam
B6
Putnam 2016 B6
Putnam 2016 B6
Source:
December 4, 2016
Putnam
Putnam 2016
Putnam calculus
Problem Statement
Evaluate
∑
k
=
1
∞
(
−
1
)
k
−
1
k
∑
n
=
0
∞
1
k
2
n
+
1
.
\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}.
k
=
1
∑
∞
k
(
−
1
)
k
−
1
n
=
0
∑
∞
k
2
n
+
1
1
.
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