Mixing
If $sumlimits^{infty}_{n=1}|x_n|^2<infty $ then ${x_n}$ is a Cauchy sequence
$begingroup$
Let ${x_n}$ be a sequence of real numbers. How do I prove that if $sum_{n=1}^infty|x_n|^2<infty$ then ${x_n}$ is a Cauchy sequence ?
Is the converse true?
How do I prove this?
For the converse, is ${x_n}=1+frac{1}{x_n}$ working?
real-analysis sequences-and-series cauchy-sequences
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
$endgroup$
add a comment |
$begingroup$
Let ${x_n}$ be a sequence of real numbers. How do I prove that if $sum_{n=1}^infty|x_n|^2<infty$ then ${x_n}$ is a Cauchy sequence ?
Is the converse true?
How do I prove this?
For the converse, is ${x_n}=1+frac{1}{x_n}$ working?
real-analysis sequences-and-series cauchy-sequences
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
$endgroup$
$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
1
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06
add a comment |
$begingroup$
Let ${x_n}$ be a sequence of real numbers. How do I prove that if $sum_{n=1}^infty|x_n|^2<infty$ then ${x_n}$ is a Cauchy sequence ?
Is the converse true?
How do I prove this?
For the converse, is ${x_n}=1+frac{1}{x_n}$ working?
real-analysis sequences-and-series cauchy-sequences
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
$endgroup$
Let ${x_n}$ be a sequence of real numbers. How do I prove that if $sum_{n=1}^infty|x_n|^2<infty$ then ${x_n}$ is a Cauchy sequence ?
Is the converse true?
How do I prove this?
For the converse, is ${x_n}=1+frac{1}{x_n}$ working?
real-analysis sequences-and-series cauchy-sequences
real-analysis sequences-and-series cauchy-sequences
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
edited Feb 11 at 8:26
Martin Sleziak
45k10123277
45k10123277
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
asked Feb 3 at 9:50
Inverse ProblemInverse Problem
1,037918
1,037918
$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
1
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06
add a comment |
$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
1
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06
$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
1
1
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06
add a comment |
1 Answer
1
active
oldest
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$begingroup$
If $displaystylesum_{n=1}^{+infty}vert x_nvert^2<+infty$, then $lim lvert x_nrvert^2=0$, hence $lim x_n=0$, so ${x_n}$ is convergent, hence it's a Cauchy sequence.
If $x_n=dfrac 1{sqrt{n}}$ for all $n>0$ then ${x_n}$ is convergent to $0$, hence is a Cauchy sequence, but $displaystylesum_{n=1}^{+infty}lvert x_nrvert^2=+infty$ (it's the harmonic series). So the converse is false.
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
$endgroup$
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $displaystylesum_{n=1}^{+infty}vert x_nvert^2<+infty$, then $lim lvert x_nrvert^2=0$, hence $lim x_n=0$, so ${x_n}$ is convergent, hence it's a Cauchy sequence.
If $x_n=dfrac 1{sqrt{n}}$ for all $n>0$ then ${x_n}$ is convergent to $0$, hence is a Cauchy sequence, but $displaystylesum_{n=1}^{+infty}lvert x_nrvert^2=+infty$ (it's the harmonic series). So the converse is false.
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
$endgroup$
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
add a comment |
$begingroup$
If $displaystylesum_{n=1}^{+infty}vert x_nvert^2<+infty$, then $lim lvert x_nrvert^2=0$, hence $lim x_n=0$, so ${x_n}$ is convergent, hence it's a Cauchy sequence.
If $x_n=dfrac 1{sqrt{n}}$ for all $n>0$ then ${x_n}$ is convergent to $0$, hence is a Cauchy sequence, but $displaystylesum_{n=1}^{+infty}lvert x_nrvert^2=+infty$ (it's the harmonic series). So the converse is false.
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
$endgroup$
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
add a comment |
$begingroup$
If $displaystylesum_{n=1}^{+infty}vert x_nvert^2<+infty$, then $lim lvert x_nrvert^2=0$, hence $lim x_n=0$, so ${x_n}$ is convergent, hence it's a Cauchy sequence.
If $x_n=dfrac 1{sqrt{n}}$ for all $n>0$ then ${x_n}$ is convergent to $0$, hence is a Cauchy sequence, but $displaystylesum_{n=1}^{+infty}lvert x_nrvert^2=+infty$ (it's the harmonic series). So the converse is false.
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
$endgroup$
If $displaystylesum_{n=1}^{+infty}vert x_nvert^2<+infty$, then $lim lvert x_nrvert^2=0$, hence $lim x_n=0$, so ${x_n}$ is convergent, hence it's a Cauchy sequence.
If $x_n=dfrac 1{sqrt{n}}$ for all $n>0$ then ${x_n}$ is convergent to $0$, hence is a Cauchy sequence, but $displaystylesum_{n=1}^{+infty}lvert x_nrvert^2=+infty$ (it's the harmonic series). So the converse is false.
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
edited Feb 3 at 10:10
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
answered Feb 3 at 9:58
Philippe MalotPhilippe Malot
2,286824
2,286824
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
add a comment |
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
....how to prove first part
$endgroup$
– Inverse Problem
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
How do I prove that its limit is $0$ or that it's a Cauchy sequence?
$endgroup$
– Philippe Malot
Feb 3 at 10:04
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
@phiiippe Malot ..i mean that sum ^{infty}_{n=1}|x_n|^2<infty then how we prove the sequence ${x_n}$ is cauchy
$endgroup$
– Inverse Problem
Feb 3 at 10:06
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
$begingroup$
Oh, that's easy. If a series is convergent, then its general term tends to $0$, in this case $lvert x_nrvert^2$ tends to $0$, hence $x_n$ tends to $0$. A convergent sequence is Cauchy.
$endgroup$
– Philippe Malot
Feb 3 at 10:08
add a comment |
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$begingroup$
Converse is not true. ${x_n} = 1+ frac{1}{x_n}$ is a counterexample.
$endgroup$
– Anthony Ter
Feb 3 at 9:59
1
$begingroup$
if $sum ^{infty}_{n=1}|x_n|^2<infty$ then $|x_n|^2to 0$ and therefore $x_nto 0$ and it is a Cauchy sequence. See en.wikipedia.org/wiki/Convergence_tests#Limit_of_the_summand
$endgroup$
– Robert Z
Feb 3 at 10:06