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Bulgaria EGMO TST
2019 Bulgaria EGMO TST
2
Hardcore algebra sequence
Hardcore algebra sequence
Source: Bulgaria EGMO TST 2019 Problem 2
December 8, 2022
algebra
Sequence
Sums and Products
series
Problem Statement
The sequence of real numbers
(
a
n
)
n
≥
0
(a_n)_{n\geq 0}
(
a
n
)
n
≥
0
is such that
a
0
=
1
a_0 = 1
a
0
=
1
,
a
1
=
a
>
2
a_1 = a > 2
a
1
=
a
>
2
and
a
n
+
1
=
(
(
a
n
a
n
−
1
)
2
−
2
)
a
n
\displaystyle a_{n+1} = \left(\left(\frac{a_n}{a_{n-1}}\right)^2 -2\right)a_n
a
n
+
1
=
(
(
a
n
−
1
a
n
)
2
−
2
)
a
n
for every positive integer
n
n
n
. Prove that
∑
i
=
0
k
1
a
i
<
2
+
a
−
a
2
−
4
2
\displaystyle \sum_{i=0}^k \frac{1}{a_i} < \frac{2+a-\sqrt{a^2-4}}{2}
i
=
0
∑
k
a
i
1
<
2
2
+
a
−
a
2
−
4
for every positive integer
k
k
k
.
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