A question on the sums of finite subsequences of a sequence of positive reals
Let $(a_n)_{ngeq 1}$ be a sequence of positive real numbers such that $a_nrightarrow 0$ and $sumlimits_{i=1}^{infty}a_i$ is divergent. Prove that the set containing the sums of all $finite$ subsequences of $a_n$ is dense in $[0,infty)$.
I know that a similar result holds. More exactly, under the same hypothesis, it was proved that the set containing the sums of $all$ subsequences (so, besides the finite subsequences, we also have the infinite ones) of $a_n$ is actually the interval $[0,infty)$. Any hint/suggestion is appreciated.
sequences-and-series
asked Dec 9 '18 at 20:02
Muffin
153
add a comment |
Let $(a_n)_{ngeq 1}$ be a sequence of positive real numbers such that $a_nrightarrow 0$ and $sumlimits_{i=1}^{infty}a_i$ is divergent. Prove that the set containing the sums of all $finite$ subsequences of $a_n$ is dense in $[0,infty)$.
I know that a similar result holds. More exactly, under the same hypothesis, it was proved that the set containing the sums of $all$ subsequences (so, besides the finite subsequences, we also have the infinite ones) of $a_n$ is actually the interval $[0,infty)$. Any hint/suggestion is appreciated.
sequences-and-series
asked Dec 9 '18 at 20:02
Muffin
153
add a comment |
Let $(a_n)_{ngeq 1}$ be a sequence of positive real numbers such that $a_nrightarrow 0$ and $sumlimits_{i=1}^{infty}a_i$ is divergent. Prove that the set containing the sums of all $finite$ subsequences of $a_n$ is dense in $[0,infty)$.
I know that a similar result holds. More exactly, under the same hypothesis, it was proved that the set containing the sums of $all$ subsequences (so, besides the finite subsequences, we also have the infinite ones) of $a_n$ is actually the interval $[0,infty)$. Any hint/suggestion is appreciated.
sequences-and-series
asked Dec 9 '18 at 20:02
Muffin
153
Let $(a_n)_{ngeq 1}$ be a sequence of positive real numbers such that $a_nrightarrow 0$ and $sumlimits_{i=1}^{infty}a_i$ is divergent. Prove that the set containing the sums of all $finite$ subsequences of $a_n$ is dense in $[0,infty)$.
I know that a similar result holds. More exactly, under the same hypothesis, it was proved that the set containing the sums of $all$ subsequences (so, besides the finite subsequences, we also have the infinite ones) of $a_n$ is actually the interval $[0,infty)$. Any hint/suggestion is appreciated.
sequences-and-series
sequences-and-series
asked Dec 9 '18 at 20:02
Muffin
153
asked Dec 9 '18 at 20:02
Muffin
153
asked Dec 9 '18 at 20:02
Muffin
153
asked Dec 9 '18 at 20:02
Muffin
153
asked Dec 9 '18 at 20:02
Muffin
153
153
add a comment |
add a comment |
2 Answers
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Well, you have almost everything you need. Let $xin[0,infty)$ and let $epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-epsilon, x+epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $sum a_{n_k}=xin(x-epsilon, x+epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $sum_{k=1}^infty a_{n_k}=x$. An equivalent way to write it is $lim_{Mtoinfty}sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $sum_{k=1}^M a_{n_k}in(x-epsilon,x+epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
answered Dec 9 '18 at 20:14
Mark
6,015415
add a comment |
You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?
Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-epsilon,r)$ for any $epsilon>0$.
answered Dec 9 '18 at 20:14
GenericMathematician
863
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
add a comment |
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2 Answers
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2 Answers
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Well, you have almost everything you need. Let $xin[0,infty)$ and let $epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-epsilon, x+epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $sum a_{n_k}=xin(x-epsilon, x+epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $sum_{k=1}^infty a_{n_k}=x$. An equivalent way to write it is $lim_{Mtoinfty}sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $sum_{k=1}^M a_{n_k}in(x-epsilon,x+epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
answered Dec 9 '18 at 20:14
Mark
6,015415
add a comment |
Well, you have almost everything you need. Let $xin[0,infty)$ and let $epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-epsilon, x+epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $sum a_{n_k}=xin(x-epsilon, x+epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $sum_{k=1}^infty a_{n_k}=x$. An equivalent way to write it is $lim_{Mtoinfty}sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $sum_{k=1}^M a_{n_k}in(x-epsilon,x+epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
answered Dec 9 '18 at 20:14
Mark
6,015415
add a comment |
Well, you have almost everything you need. Let $xin[0,infty)$ and let $epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-epsilon, x+epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $sum a_{n_k}=xin(x-epsilon, x+epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $sum_{k=1}^infty a_{n_k}=x$. An equivalent way to write it is $lim_{Mtoinfty}sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $sum_{k=1}^M a_{n_k}in(x-epsilon,x+epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
answered Dec 9 '18 at 20:14
Mark
6,015415
Well, you have almost everything you need. Let $xin[0,infty)$ and let $epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-epsilon, x+epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $sum a_{n_k}=xin(x-epsilon, x+epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $sum_{k=1}^infty a_{n_k}=x$. An equivalent way to write it is $lim_{Mtoinfty}sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $sum_{k=1}^M a_{n_k}in(x-epsilon,x+epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
answered Dec 9 '18 at 20:14
Mark
6,015415
answered Dec 9 '18 at 20:14
Mark
6,015415
answered Dec 9 '18 at 20:14
Mark
6,015415
answered Dec 9 '18 at 20:14
Mark
6,015415
6,015415
add a comment |
add a comment |
You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?
Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-epsilon,r)$ for any $epsilon>0$.
answered Dec 9 '18 at 20:14
GenericMathematician
863
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
add a comment |
You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?
Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-epsilon,r)$ for any $epsilon>0$.
answered Dec 9 '18 at 20:14
GenericMathematician
863
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
add a comment |
You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?
Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-epsilon,r)$ for any $epsilon>0$.
answered Dec 9 '18 at 20:14
GenericMathematician
863
You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?
Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-epsilon,r)$ for any $epsilon>0$.
answered Dec 9 '18 at 20:14
GenericMathematician
863
answered Dec 9 '18 at 20:14
GenericMathematician
863
answered Dec 9 '18 at 20:14
GenericMathematician
863
answered Dec 9 '18 at 20:14
GenericMathematician
863
863
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
add a comment |
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,infty)$.
– Muffin
Dec 9 '18 at 20:32
add a comment |
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