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Lebl-Basic-Analysis-I
· Problem 2.1.16
Sequences and Series
Mathematics · sequences-and-limits
Problem
Let
{
x
n
}
n
=
1
∞
\{x_n\}_{n=1}^\infty
{
x
n
}
n
=
1
∞
be a sequence. Suppose there are two convergent subsequences
{
x
n
i
}
i
=
1
∞
\{x_{n_i}\}_{i=1}^\infty
{
x
n
i
}
i
=
1
∞
and
{
x
m
i
}
i
=
1
∞
\{x_{m_i}\}_{i=1}^\infty
{
x
m
i
}
i
=
1
∞
. Suppose
lim
i
→
∞
x
n
i
=
a
and
lim
i
→
∞
x
m
i
=
b
,
\lim_{i\to\infty} x_{n_i} = a \quad \text{and} \quad \lim_{i\to\infty} x_{m_i} = b,
i
→
∞
lim
x
n
i
=
a
and
i
→
∞
lim
x
m
i
=
b
,
where
a
≠
b
a \neq b
a
=
b
. Prove that
{
x
n
}
n
=
1
∞
\{x_n\}_{n=1}^\infty
{
x
n
}
n
=
1
∞
is not convergent, without using Proposition 2.1.17.
Related theory
sequences-and-limits
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