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Putnam
2015 Putnam
B6
Putnam 2015 B6
Putnam 2015 B6
Source:
December 6, 2015
Putnam
Putnam 2015
Putnam number theory
Problem Statement
For each positive integer
k
,
k,
k
,
let
A
(
k
)
A(k)
A
(
k
)
be the number of odd divisors of
k
k
k
in the interval
[
1
,
2
k
)
.
\left[1,\sqrt{2k}\right).
[
1
,
2
k
)
.
Evaluate:
∑
k
=
1
∞
(
−
1
)
k
−
1
A
(
k
)
k
.
\sum_{k=1}^{\infty}(-1)^{k-1}\frac{A(k)}k.
k
=
1
∑
∞
(
−
1
)
k
−
1
k
A
(
k
)
.
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