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2016 Brazil Undergrad MO
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1
Part of
2016 Brazil Undergrad MO
Problems
(1)
'Amortized' series implies limit of arithmetical mean = 0
Source: 38th Brazilian Undergrad MO (2016) - First Day, Problem 1
11/25/2016
Let
(
a
n
)
n
≥
1
(a_n)_{n \geq 1}
(
a
n
)
n
≥
1
s sequence of reals such that
∑
n
≥
1
a
n
n
\sum_{n \geq 1}{\frac{a_n}{n}}
n
≥
1
∑
n
a
n
converges. Show that
lim
n
→
∞
1
n
⋅
∑
k
=
1
n
a
k
=
0
\lim_{n \rightarrow \infty}{\frac{1}{n} \cdot \sum_{k=1}^{n}{a_k}} = 0
n
→
∞
lim
n
1
⋅
k
=
1
∑
n
a
k
=
0
real analysis