every real in (0,1) written as linear combination of the terms of a sequence
$begingroup$
Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?
real-analysis
asked Dec 1 '18 at 15:02
LanceLance
8912
$endgroup$
add a comment |
$begingroup$
Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?
real-analysis
asked Dec 1 '18 at 15:02
LanceLance
8912
$endgroup$
1
$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28
$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33
$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41
add a comment |
$begingroup$
Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?
real-analysis
asked Dec 1 '18 at 15:02
LanceLance
8912
$endgroup$
Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?
real-analysis
real-analysis
asked Dec 1 '18 at 15:02
LanceLance
8912
asked Dec 1 '18 at 15:02
LanceLance
8912
edited Dec 1 '18 at 16:53
Lance
asked Dec 1 '18 at 15:02
LanceLance
8912
asked Dec 1 '18 at 15:02
LanceLance
8912
asked Dec 1 '18 at 15:02
LanceLance
8912
8912
1
$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28
$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33
$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41
add a comment |
1
$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28
$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33
$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41
1
1
$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28
$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28
$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33
$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33
$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41
$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).
However, we can show that the following are equivalent:
(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.
(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.
(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.
$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.
Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.
$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.
This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.
(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.
Remark:
If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
$endgroup$
add a comment |
$begingroup$
But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
By naturals I mean $1,2...$
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
add a comment |
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2 Answers
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$begingroup$
If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).
However, we can show that the following are equivalent:
(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.
(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.
(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.
$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.
Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.
$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.
This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.
(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.
Remark:
If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
$endgroup$
add a comment |
$begingroup$
If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).
However, we can show that the following are equivalent:
(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.
(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.
(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.
$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.
Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.
$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.
This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.
(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.
Remark:
If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
$endgroup$
add a comment |
$begingroup$
If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).
However, we can show that the following are equivalent:
(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.
(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.
(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.
$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.
Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.
$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.
This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.
(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.
Remark:
If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
$endgroup$
If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).
However, we can show that the following are equivalent:
(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.
(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.
(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.
$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.
Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.
$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.
This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.
(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.
Remark:
If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
edited Dec 2 '18 at 9:27
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
answered Dec 1 '18 at 19:09
Paul FrostPaul Frost
9,9153932
9,9153932
add a comment |
add a comment |
$begingroup$
But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
By naturals I mean $1,2...$
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
add a comment |
$begingroup$
But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
By naturals I mean $1,2...$
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
add a comment |
$begingroup$
But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
By naturals I mean $1,2...$
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
By naturals I mean $1,2...$
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:54
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
6,542917
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
add a comment |
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
$begingroup$
yes I forgot to add it. Anyway, the weights can be also 0
$endgroup$
– Lance
Dec 1 '18 at 16:57
add a comment |
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It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
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– lulu
Dec 1 '18 at 15:28
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In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
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– Paul Frost
Dec 1 '18 at 17:33
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lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
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– Paul Frost
Dec 1 '18 at 17:41