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IMC
2010 IMC
2
2
Part of
2010 IMC
Problems
(2)
IMC 2010 - Problem 2
Source:
7/26/2010
Compute the sum of the series
∑
k
=
0
∞
1
(
4
k
+
1
)
(
4
k
+
2
)
(
4
k
+
3
)
(
4
k
+
4
)
=
1
1
⋅
2
⋅
3
⋅
4
+
1
5
⋅
6
⋅
7
⋅
8
+
.
.
.
\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{1\cdot2\cdot3\cdot4} + \frac{1}{5\cdot6\cdot7\cdot8} + ...
∑
k
=
0
∞
(
4
k
+
1
)
(
4
k
+
2
)
(
4
k
+
3
)
(
4
k
+
4
)
1
=
1
⋅
2
⋅
3
⋅
4
1
+
5
⋅
6
⋅
7
⋅
8
1
+
...
integration
logarithms
real analysis
infinite series
IMC 2010, Problem 2, Day 2
Source:
7/27/2010
Let
a
0
,
a
1
,
…
,
a
n
a_0,a_1,\dots,a_n
a
0
,
a
1
,
…
,
a
n
be positive real numbers such that
a
k
+
1
−
a
k
≥
1
a_{k+1}-a_k \geq 1
a
k
+
1
−
a
k
≥
1
for all
k
=
0
,
1
,
…
,
n
−
1.
k=0,1,\dots,n-1.
k
=
0
,
1
,
…
,
n
−
1.
Prove that
1
+
1
a
0
(
1
+
1
a
1
−
a
0
)
⋯
(
1
+
1
a
n
−
a
0
)
≤
(
1
+
1
a
0
)
(
1
+
1
a
1
)
⋯
(
1
+
1
a
n
)
.
1+\frac{1}{a_0} \left( 1+\frac1{a_1-a_0}\right)\cdots\left(1+\frac1{a_n-a_0}\right)\leq \left(1+\frac1{a_0}\right) \left(1+\frac1{a_1}\right)\cdots \left(1+\frac1{a_n}\right).
1
+
a
0
1
(
1
+
a
1
−
a
0
1
)
⋯
(
1
+
a
n
−
a
0
1
)
≤
(
1
+
a
0
1
)
(
1
+
a
1
1
)
⋯
(
1
+
a
n
1
)
.
inequalities
induction
inequalities unsolved