Mixing
sum of sequences converging to $0$
$begingroup$
Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:
- For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.
- For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.
Is it true that:
$$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
}lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$
Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$
sequences-and-series analysis convergence local-field
asked Jan 23 at 19:17
manifoldmanifold
302213
$endgroup$
add a comment |
$begingroup$
Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:
- For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.
- For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.
Is it true that:
$$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
}lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$
Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$
sequences-and-series analysis convergence local-field
asked Jan 23 at 19:17
manifoldmanifold
302213
$endgroup$
add a comment |
$begingroup$
Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:
- For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.
- For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.
Is it true that:
$$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
}lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$
Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$
sequences-and-series analysis convergence local-field
asked Jan 23 at 19:17
manifoldmanifold
302213
$endgroup$
Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:
- For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.
- For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.
Is it true that:
$$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
}lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$
Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$
sequences-and-series analysis convergence local-field
sequences-and-series analysis convergence local-field
asked Jan 23 at 19:17
manifoldmanifold
302213
asked Jan 23 at 19:17
manifoldmanifold
302213
edited Jan 23 at 19:24
manifold
asked Jan 23 at 19:17
manifoldmanifold
302213
asked Jan 23 at 19:17
manifoldmanifold
302213
asked Jan 23 at 19:17
manifoldmanifold
302213
302213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider a list of sequences:
$$
a_1^{(n)}=1,0,0,0,0,dots
$$
$$
a_2^{(n)}=0,1,0,0,0,dots
$$
$$
a_3^{(n)}=0,0,1,0,0,dots
$$
$$
a_4^{(n)}=0,0,0,1,0,dots
$$
$$
vdots
$$
Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.
answered Jan 23 at 19:33
ZeeklessZeekless
577111
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider a list of sequences:
$$
a_1^{(n)}=1,0,0,0,0,dots
$$
$$
a_2^{(n)}=0,1,0,0,0,dots
$$
$$
a_3^{(n)}=0,0,1,0,0,dots
$$
$$
a_4^{(n)}=0,0,0,1,0,dots
$$
$$
vdots
$$
Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.
answered Jan 23 at 19:33
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Consider a list of sequences:
$$
a_1^{(n)}=1,0,0,0,0,dots
$$
$$
a_2^{(n)}=0,1,0,0,0,dots
$$
$$
a_3^{(n)}=0,0,1,0,0,dots
$$
$$
a_4^{(n)}=0,0,0,1,0,dots
$$
$$
vdots
$$
Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.
answered Jan 23 at 19:33
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Consider a list of sequences:
$$
a_1^{(n)}=1,0,0,0,0,dots
$$
$$
a_2^{(n)}=0,1,0,0,0,dots
$$
$$
a_3^{(n)}=0,0,1,0,0,dots
$$
$$
a_4^{(n)}=0,0,0,1,0,dots
$$
$$
vdots
$$
Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.
answered Jan 23 at 19:33
ZeeklessZeekless
577111
$endgroup$
Consider a list of sequences:
$$
a_1^{(n)}=1,0,0,0,0,dots
$$
$$
a_2^{(n)}=0,1,0,0,0,dots
$$
$$
a_3^{(n)}=0,0,1,0,0,dots
$$
$$
a_4^{(n)}=0,0,0,1,0,dots
$$
$$
vdots
$$
Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.
answered Jan 23 at 19:33
ZeeklessZeekless
577111
answered Jan 23 at 19:33
ZeeklessZeekless
577111
answered Jan 23 at 19:33
ZeeklessZeekless
577111
answered Jan 23 at 19:33
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
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