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1990 IMO Longlists
35
Sum of series - ILL 1990 Iran 4
Sum of series - ILL 1990 Iran 4
Source:
September 18, 2010
calculus
derivative
geometric series
algebra unsolved
algebra
Problem Statement
Prove that if
∣
x
∣
<
1
|x| < 1
∣
x
∣
<
1
, then
x
(
1
−
x
)
2
+
x
2
(
1
+
x
2
)
2
+
x
3
(
1
−
x
3
)
2
+
⋯
=
x
1
−
x
+
2
x
2
1
+
x
2
+
3
x
3
1
−
x
3
+
⋯
\frac{x}{(1-x)^2}+\frac{x^2}{(1+x^2)^2} + \frac{x^3}{(1-x^3)^2}+\cdots=\frac{x}{1-x}+\frac{2x^2}{1+x^2}+\frac{3x^3}{1-x^3}+\cdots
(
1
−
x
)
2
x
+
(
1
+
x
2
)
2
x
2
+
(
1
−
x
3
)
2
x
3
+
⋯
=
1
−
x
x
+
1
+
x
2
2
x
2
+
1
−
x
3
3
x
3
+
⋯
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