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Today's Calculation Of Integral
2007 Today's Calculation Of Integral
175
Today's calculation of Integral 175
Today's calculation of Integral 175
Source: Kwansei Gakuin University entrance exam 2007
February 4, 2007
calculus
integration
logarithms
calculus computations
Problem Statement
Evaluate
∑
n
=
0
∞
1
(
2
n
+
1
)
2
2
n
+
1
.
\sum_{n=0}^{\infty}\frac{1}{(2n+1)2^{2n+1}}.
∑
n
=
0
∞
(
2
n
+
1
)
2
2
n
+
1
1
.
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