Mixing
$sum_{n}^{infty}u_n$ converges $Rightarrow $ $sum_{n}^{infty }nu_n^2$ and $sum_{n}^{infty}frac{u_n}{1-nu_n}$...
$begingroup$
Let $(u_n)_{ngeq 1}$ be a sequence with $u_nin mathbb{R}_+$, such that $sum_{n}^{infty}u_n$ converges.
a)Suppose that $(u_n)_{ngeq 1}$ decreases:
1,Prove that $nu_nrightarrow 0$ as $nrightarrow infty$.
2,Do the series $sum_{n}^{infty }nu_n^2$ and $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverge?
b) Examine 2, when not supposing that $(u_n)_{ngeq 1}$ decreases.
I'm not sure about my solution for a)
We have $0leq frac{n}{2}u_nleq sum_{k=left lfloor frac{n}{2} right rfloor}^{n}u_krightarrow 0$, $Rightarrow nu_nrightarrow 0 $ as $nrightarrow infty$.
$exists Minmathbb{R}_+:forall ninmathbb{N}^*, nu_nleq M$
$Rightarrow 0leq nu_n^2leq Mu_n$, hence: $sum_{n}^{infty }nu_n^2$ converges.
$frac{u_n}{1-nu_n}sim u_n as nrightarrow infty$, hence: $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverges.
I can't solve b). Could you give me some hints?
real-analysis sequences-and-series
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
$endgroup$
add a comment |
$begingroup$
Let $(u_n)_{ngeq 1}$ be a sequence with $u_nin mathbb{R}_+$, such that $sum_{n}^{infty}u_n$ converges.
a)Suppose that $(u_n)_{ngeq 1}$ decreases:
1,Prove that $nu_nrightarrow 0$ as $nrightarrow infty$.
2,Do the series $sum_{n}^{infty }nu_n^2$ and $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverge?
b) Examine 2, when not supposing that $(u_n)_{ngeq 1}$ decreases.
I'm not sure about my solution for a)
We have $0leq frac{n}{2}u_nleq sum_{k=left lfloor frac{n}{2} right rfloor}^{n}u_krightarrow 0$, $Rightarrow nu_nrightarrow 0 $ as $nrightarrow infty$.
$exists Minmathbb{R}_+:forall ninmathbb{N}^*, nu_nleq M$
$Rightarrow 0leq nu_n^2leq Mu_n$, hence: $sum_{n}^{infty }nu_n^2$ converges.
$frac{u_n}{1-nu_n}sim u_n as nrightarrow infty$, hence: $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverges.
I can't solve b). Could you give me some hints?
real-analysis sequences-and-series
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
$endgroup$
1
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
$endgroup$
– reuns
Sep 17 '16 at 20:05
$begingroup$
For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:27
$begingroup$
The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:34
add a comment |
$begingroup$
Let $(u_n)_{ngeq 1}$ be a sequence with $u_nin mathbb{R}_+$, such that $sum_{n}^{infty}u_n$ converges.
a)Suppose that $(u_n)_{ngeq 1}$ decreases:
1,Prove that $nu_nrightarrow 0$ as $nrightarrow infty$.
2,Do the series $sum_{n}^{infty }nu_n^2$ and $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverge?
b) Examine 2, when not supposing that $(u_n)_{ngeq 1}$ decreases.
I'm not sure about my solution for a)
We have $0leq frac{n}{2}u_nleq sum_{k=left lfloor frac{n}{2} right rfloor}^{n}u_krightarrow 0$, $Rightarrow nu_nrightarrow 0 $ as $nrightarrow infty$.
$exists Minmathbb{R}_+:forall ninmathbb{N}^*, nu_nleq M$
$Rightarrow 0leq nu_n^2leq Mu_n$, hence: $sum_{n}^{infty }nu_n^2$ converges.
$frac{u_n}{1-nu_n}sim u_n as nrightarrow infty$, hence: $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverges.
I can't solve b). Could you give me some hints?
real-analysis sequences-and-series
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
$endgroup$
Let $(u_n)_{ngeq 1}$ be a sequence with $u_nin mathbb{R}_+$, such that $sum_{n}^{infty}u_n$ converges.
a)Suppose that $(u_n)_{ngeq 1}$ decreases:
1,Prove that $nu_nrightarrow 0$ as $nrightarrow infty$.
2,Do the series $sum_{n}^{infty }nu_n^2$ and $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverge?
b) Examine 2, when not supposing that $(u_n)_{ngeq 1}$ decreases.
I'm not sure about my solution for a)
We have $0leq frac{n}{2}u_nleq sum_{k=left lfloor frac{n}{2} right rfloor}^{n}u_krightarrow 0$, $Rightarrow nu_nrightarrow 0 $ as $nrightarrow infty$.
$exists Minmathbb{R}_+:forall ninmathbb{N}^*, nu_nleq M$
$Rightarrow 0leq nu_n^2leq Mu_n$, hence: $sum_{n}^{infty }nu_n^2$ converges.
$frac{u_n}{1-nu_n}sim u_n as nrightarrow infty$, hence: $sum_{n}^{infty}frac{u_n}{1-nu_n}$ coverges.
I can't solve b). Could you give me some hints?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
asked Sep 17 '16 at 19:15
Phuong VuPhuong Vu
1846
1846
1
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
$endgroup$
– reuns
Sep 17 '16 at 20:05
$begingroup$
For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:27
$begingroup$
The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:34
add a comment |
1
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
$endgroup$
– reuns
Sep 17 '16 at 20:05
$begingroup$
For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:27
$begingroup$
The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:34
1
1
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
$endgroup$
– reuns
Sep 17 '16 at 20:05
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
$endgroup$
– reuns
Sep 17 '16 at 20:05
$begingroup$
For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:27
$begingroup$
For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:27
$begingroup$
The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:34
$begingroup$
The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
$endgroup$
– DanielWainfleet
Jan 29 at 6:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The $sum_n^infty u_n= sum_n^infty nu_n frac{1}{n}$, if $n u_n to k>0$ then
the sum is like $sum_n^infty frac{k}{n}= infty$, and i didn't use the decrease argument.
If $sum_{n=p}^infty nu_n^2le sum_{n=p}^infty u_n< infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- epsilon$ for all $ nge p$ then $sum_{n=p}^infty frac{u_n}{1- nu_n}le sum_{n=p}^infty frac{u_n}{1-epsilon}$, hence 2) is still valid without the decrease.
It is not valid if we eliminate the $u_n in mathbb{R}_+$ hiypothesis.
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
$endgroup$
add a comment |
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$begingroup$
The $sum_n^infty u_n= sum_n^infty nu_n frac{1}{n}$, if $n u_n to k>0$ then
the sum is like $sum_n^infty frac{k}{n}= infty$, and i didn't use the decrease argument.
If $sum_{n=p}^infty nu_n^2le sum_{n=p}^infty u_n< infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- epsilon$ for all $ nge p$ then $sum_{n=p}^infty frac{u_n}{1- nu_n}le sum_{n=p}^infty frac{u_n}{1-epsilon}$, hence 2) is still valid without the decrease.
It is not valid if we eliminate the $u_n in mathbb{R}_+$ hiypothesis.
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
$endgroup$
add a comment |
$begingroup$
The $sum_n^infty u_n= sum_n^infty nu_n frac{1}{n}$, if $n u_n to k>0$ then
the sum is like $sum_n^infty frac{k}{n}= infty$, and i didn't use the decrease argument.
If $sum_{n=p}^infty nu_n^2le sum_{n=p}^infty u_n< infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- epsilon$ for all $ nge p$ then $sum_{n=p}^infty frac{u_n}{1- nu_n}le sum_{n=p}^infty frac{u_n}{1-epsilon}$, hence 2) is still valid without the decrease.
It is not valid if we eliminate the $u_n in mathbb{R}_+$ hiypothesis.
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
$endgroup$
add a comment |
$begingroup$
The $sum_n^infty u_n= sum_n^infty nu_n frac{1}{n}$, if $n u_n to k>0$ then
the sum is like $sum_n^infty frac{k}{n}= infty$, and i didn't use the decrease argument.
If $sum_{n=p}^infty nu_n^2le sum_{n=p}^infty u_n< infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- epsilon$ for all $ nge p$ then $sum_{n=p}^infty frac{u_n}{1- nu_n}le sum_{n=p}^infty frac{u_n}{1-epsilon}$, hence 2) is still valid without the decrease.
It is not valid if we eliminate the $u_n in mathbb{R}_+$ hiypothesis.
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
$endgroup$
The $sum_n^infty u_n= sum_n^infty nu_n frac{1}{n}$, if $n u_n to k>0$ then
the sum is like $sum_n^infty frac{k}{n}= infty$, and i didn't use the decrease argument.
If $sum_{n=p}^infty nu_n^2le sum_{n=p}^infty u_n< infty$ if $p$ is large enough, and if $p$ is such that $1 - nu_n> 1- epsilon$ for all $ nge p$ then $sum_{n=p}^infty frac{u_n}{1- nu_n}le sum_{n=p}^infty frac{u_n}{1-epsilon}$, hence 2) is still valid without the decrease.
It is not valid if we eliminate the $u_n in mathbb{R}_+$ hiypothesis.
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
answered Jan 28 at 23:11
Claudio DelfinoClaudio Delfino
63
63
add a comment |
add a comment |
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1
$begingroup$
The answer to $a$ is correct. Good job
$endgroup$
– b00n heT
Sep 17 '16 at 19:25
$begingroup$
for those exercices, in general you have to use the partial summation $sum_{n=1}^N a_n u_n = v(N)a_N +sum_{n=0}^{N -1} v(n)(a_n-a_{n+1})$ where $v(N) = sum_{n=1}^N u_n$. since $sum_n u_n$ converges, $|v(n)| < C$, and with $a_n = u_n n$, you have $sum_{n=0}^{N -1} v(n)(a_n-a_{n+1}) = sum_{n=0}^{N -1} v(n)(u_n n-u_{n+1} n) < sum_{n=0}^{N -1} C|u_n|$. So it reduces to $v(N)u_N N to 0$, i.e. $u_n = o(1/n)$, which is the case since otherwise ($u_n$ being decreasing) $sum_n u_n$ wouldn't converge
$endgroup$
– reuns
Sep 17 '16 at 20:03
$begingroup$
and since $u_n = o(1/n)$, you have $n = o(1/u_n)$ and $frac{u_n}{1-n u_n} =frac{1}{ frac{1}{u_n}-n} sim u_n$ so yes $sum_n frac{u_n}{1-n u_n} $ converges (for b) note how $u_n = o(1/n)$ was important)
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– reuns
Sep 17 '16 at 20:05
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For (b) suppose $a_{(4^n)}=1/2^n$ and $a_m=1/2^m$ when $m$ is not a power of $4.$ Then $sum_{jin Bbb N} a_j<sum_{n in Bbb N}1/2^n+sum_{min Bbb N}1/2^m=2$. But $4^n(a_{(4^n)})^2=1.$
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– DanielWainfleet
Jan 29 at 6:27
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The idea in my previous comment is that we can have $a_j=1/sqrt j$ for a "sparse" but infinite collection of $j$'s and still have $sum_na_n<infty.$
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– DanielWainfleet
Jan 29 at 6:34