If $limsup fraca_n+1a_n<1$ does $sum a_n$ converge even if $limfraca_n+1a_n$ does not exist?Does $sumfracsin nn$ converge?Suppose $a_n geq 0$, and $sum a_n$ diverges, and $lim a_n = 0$. Show that $sum fraca_n1+a_n$ diverges.Proving $limsupfrac 1 a_n=frac 1 liminf a_n$ and $limsup a_ncdot limsup frac 1 a_n ge 1$Does $sum_1^infty fracnn^2 + 4$ converge or diverge? Is my solution correct?Suppose $suma_n$ converges. How do I prove that $sumfracsqrta_nn$ convergesGiven that $lim _n to infty a_n = 0$, does $displaystyle sum _n=1^infty |na_n|$ converge?prove convergence: $limsup a_n/b_n < ∞$, $sum b_n$ convergent implies $sum a_n$ convergentIf $a_n$ is sequence of positive numbers such that $sum a_n$ converges, then does $sum frac (a_n)^frac 14n^frac 45$ converge?Convergence of $sumlimits_n=1^inftya_n$ implies convergence of $sumlimits_n=1^inftya_n^sigma_n$ where $sigma_n=fracnn+1$?Does $fraca_n+1a_n < (fracnn+1)^2$ imply $sum a_n <infty$ here?
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If $limsup fraca_n+1a_n
Does $sumfracsin nn$ converge?Suppose $a_n geq 0$, and $sum a_n$ diverges, and $lim a_n = 0$. Show that $sum fraca_n1+a_n$ diverges.Proving $limsupfrac 1 a_n=frac 1 liminf a_n$ and $limsup a_ncdot limsup frac 1 a_n ge 1$Does $sum_1^infty fracnn^2 + 4$ converge or diverge? Is my solution correct?Suppose $suma_n$ converges. How do I prove that $sumfracsqrta_nn$ convergesGiven that $lim _n to infty a_n = 0$, does $displaystyle sum _n=1^infty |na_n|$ converge?prove convergence: $limsup a_n/b_n < ∞$, $sum b_n$ convergent implies $sum a_n$ convergentIf $a_n$ is sequence of positive numbers such that $sum a_n$ converges, then does $sum frac (a_n)^frac 14n^frac 45$ converge?Convergence of $sumlimits_n=1^inftya_n$ implies convergence of $sumlimits_n=1^inftya_n^sigma_n$ where $sigma_n=fracnn+1$?Does $fraca_n+1a_n < (fracnn+1)^2$ imply $sum a_n <infty$ here?
$begingroup$
I was wondering why the Ratio test has the $lim$ sign and the root test the $limsup$ sign.
- Quotient test: $lim|fraca_n+1a_n|<1Rightarrow sum a_n$
converges. - Ratio test: $limsup sqrt[n]<1Rightarrow sum a_n$
converges.
If I pick
$$a_n=begincases2^-fracn2 &mboxn even\3^-fracn+12 &mboxn odd endcases,$$
then for $fraca_n+1a_n$ we either have $(3/2)^fracn+12$ or $(2/3)^fracn2cdot 3$. It is not bounded therefore the limit does not exist, and $limsup=infty$ makes no difference.
The textbook says that this example shows that a Quotient test analogously to the Ratio test with $limsup|fraca_n+1a_n|$ instead of $lim|fraca_n+1a_n|$ is not true.
What does this sentence mean why does it justify that if we have $limsup<1$ we do not have necessarily $sum a_n<infty$. Because that is what I am getting out of it.
Edit:
I have understood it now the Ratio test also says if $lim fraca_n+1a_n>1$ the series diverges. If I would change this with $limsup$ then we would get a contradiction with the root test
real-analysis sequences-and-series proof-explanation
edited Mar 22 at 16:25
Micah
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
$endgroup$
add a comment |
$begingroup$
I was wondering why the Ratio test has the $lim$ sign and the root test the $limsup$ sign.
- Quotient test: $lim|fraca_n+1a_n|<1Rightarrow sum a_n$
converges. - Ratio test: $limsup sqrt[n]<1Rightarrow sum a_n$
converges.
If I pick
$$a_n=begincases2^-fracn2 &mboxn even\3^-fracn+12 &mboxn odd endcases,$$
then for $fraca_n+1a_n$ we either have $(3/2)^fracn+12$ or $(2/3)^fracn2cdot 3$. It is not bounded therefore the limit does not exist, and $limsup=infty$ makes no difference.
The textbook says that this example shows that a Quotient test analogously to the Ratio test with $limsup|fraca_n+1a_n|$ instead of $lim|fraca_n+1a_n|$ is not true.
What does this sentence mean why does it justify that if we have $limsup<1$ we do not have necessarily $sum a_n<infty$. Because that is what I am getting out of it.
Edit:
I have understood it now the Ratio test also says if $lim fraca_n+1a_n>1$ the series diverges. If I would change this with $limsup$ then we would get a contradiction with the root test
real-analysis sequences-and-series proof-explanation
edited Mar 22 at 16:25
Micah
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
$endgroup$
$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31
add a comment |
$begingroup$
I was wondering why the Ratio test has the $lim$ sign and the root test the $limsup$ sign.
- Quotient test: $lim|fraca_n+1a_n|<1Rightarrow sum a_n$
converges. - Ratio test: $limsup sqrt[n]<1Rightarrow sum a_n$
converges.
If I pick
$$a_n=begincases2^-fracn2 &mboxn even\3^-fracn+12 &mboxn odd endcases,$$
then for $fraca_n+1a_n$ we either have $(3/2)^fracn+12$ or $(2/3)^fracn2cdot 3$. It is not bounded therefore the limit does not exist, and $limsup=infty$ makes no difference.
The textbook says that this example shows that a Quotient test analogously to the Ratio test with $limsup|fraca_n+1a_n|$ instead of $lim|fraca_n+1a_n|$ is not true.
What does this sentence mean why does it justify that if we have $limsup<1$ we do not have necessarily $sum a_n<infty$. Because that is what I am getting out of it.
Edit:
I have understood it now the Ratio test also says if $lim fraca_n+1a_n>1$ the series diverges. If I would change this with $limsup$ then we would get a contradiction with the root test
real-analysis sequences-and-series proof-explanation
edited Mar 22 at 16:25
Micah
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
$endgroup$
I was wondering why the Ratio test has the $lim$ sign and the root test the $limsup$ sign.
- Quotient test: $lim|fraca_n+1a_n|<1Rightarrow sum a_n$
converges. - Ratio test: $limsup sqrt[n]<1Rightarrow sum a_n$
converges.
If I pick
$$a_n=begincases2^-fracn2 &mboxn even\3^-fracn+12 &mboxn odd endcases,$$
then for $fraca_n+1a_n$ we either have $(3/2)^fracn+12$ or $(2/3)^fracn2cdot 3$. It is not bounded therefore the limit does not exist, and $limsup=infty$ makes no difference.
The textbook says that this example shows that a Quotient test analogously to the Ratio test with $limsup|fraca_n+1a_n|$ instead of $lim|fraca_n+1a_n|$ is not true.
What does this sentence mean why does it justify that if we have $limsup<1$ we do not have necessarily $sum a_n<infty$. Because that is what I am getting out of it.
Edit:
I have understood it now the Ratio test also says if $lim fraca_n+1a_n>1$ the series diverges. If I would change this with $limsup$ then we would get a contradiction with the root test
real-analysis sequences-and-series proof-explanation
real-analysis sequences-and-series proof-explanation
edited Mar 22 at 16:25
Micah
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
edited Mar 22 at 16:25
Micah
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
edited Mar 22 at 16:25
Micah
30.3k1364106
edited Mar 22 at 16:25
Micah
30.3k1364106
edited Mar 22 at 16:25
Micah
30.3k1364106
30.3k1364106
asked Mar 22 at 15:42
New2MathNew2Math
16715
asked Mar 22 at 15:42
New2MathNew2Math
16715
asked Mar 22 at 15:42
New2MathNew2Math
16715
16715
$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31
add a comment |
$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31
$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $limsup_nleftlvertfraca_n+1a_nrightrvert<1$, then take some $cinleft(limsup_nleftlvertfraca_n+1a_nrightrvert,1right)$. Then $limsup_nleftlvertfraca_n+1a_nrightrvert<c$ and so, for some $Ninmathbb N$, if $ngeqslant N$, then $leftlvertfraca_n+1a_nrightrvert<c$. But then $lvert a_N+1rvert<clvert a_Nrvert$, $lvert a_N+2rvert<c^2lvert a_Nrvert$ and so on. So, yes, the series $sum_n=0^infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
The ratio test can be expressed as follows.
Let $ell=liminf_nto inftyleft|fraca_n+1a_nright|$ and let $L=limsup_nto inftyleft|fraca_n+1a_nright|$. Then, the series $$sum_n=1^infty a_nbegincases
textconverges (absolutey)&, L<1\\
textdiverges &, ell>1\\
textdiverges &, left|fraca_n+1a_nright|ge1,textfor,n,textlarge\\
textinconclusive&,textotherwise
endcases$$
The root test is stronger than the ratio test since
$$liminf_ntoinftyleft|fraca_n+1a_nright|le liminf_nto inftysqrt[n]le limsup_ntoinftysqrt[n]le limsup_ntoinftyleft|fraca_n+1a_nright|$$
In the example in the OP, the ratio test in inconclusive since $L=infty$ and $ell=0$. However, the root test reveals
$$limsup_ntoinftysqrt[n]=2^-1/2<1$$
and the series converges.
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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votes
$begingroup$
If $limsup_nleftlvertfraca_n+1a_nrightrvert<1$, then take some $cinleft(limsup_nleftlvertfraca_n+1a_nrightrvert,1right)$. Then $limsup_nleftlvertfraca_n+1a_nrightrvert<c$ and so, for some $Ninmathbb N$, if $ngeqslant N$, then $leftlvertfraca_n+1a_nrightrvert<c$. But then $lvert a_N+1rvert<clvert a_Nrvert$, $lvert a_N+2rvert<c^2lvert a_Nrvert$ and so on. So, yes, the series $sum_n=0^infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
If $limsup_nleftlvertfraca_n+1a_nrightrvert<1$, then take some $cinleft(limsup_nleftlvertfraca_n+1a_nrightrvert,1right)$. Then $limsup_nleftlvertfraca_n+1a_nrightrvert<c$ and so, for some $Ninmathbb N$, if $ngeqslant N$, then $leftlvertfraca_n+1a_nrightrvert<c$. But then $lvert a_N+1rvert<clvert a_Nrvert$, $lvert a_N+2rvert<c^2lvert a_Nrvert$ and so on. So, yes, the series $sum_n=0^infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
If $limsup_nleftlvertfraca_n+1a_nrightrvert<1$, then take some $cinleft(limsup_nleftlvertfraca_n+1a_nrightrvert,1right)$. Then $limsup_nleftlvertfraca_n+1a_nrightrvert<c$ and so, for some $Ninmathbb N$, if $ngeqslant N$, then $leftlvertfraca_n+1a_nrightrvert<c$. But then $lvert a_N+1rvert<clvert a_Nrvert$, $lvert a_N+2rvert<c^2lvert a_Nrvert$ and so on. So, yes, the series $sum_n=0^infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
If $limsup_nleftlvertfraca_n+1a_nrightrvert<1$, then take some $cinleft(limsup_nleftlvertfraca_n+1a_nrightrvert,1right)$. Then $limsup_nleftlvertfraca_n+1a_nrightrvert<c$ and so, for some $Ninmathbb N$, if $ngeqslant N$, then $leftlvertfraca_n+1a_nrightrvert<c$. But then $lvert a_N+1rvert<clvert a_Nrvert$, $lvert a_N+2rvert<c^2lvert a_Nrvert$ and so on. So, yes, the series $sum_n=0^infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 22 at 15:49
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
add a comment |
add a comment |
$begingroup$
The ratio test can be expressed as follows.
Let $ell=liminf_nto inftyleft|fraca_n+1a_nright|$ and let $L=limsup_nto inftyleft|fraca_n+1a_nright|$. Then, the series $$sum_n=1^infty a_nbegincases
textconverges (absolutey)&, L<1\\
textdiverges &, ell>1\\
textdiverges &, left|fraca_n+1a_nright|ge1,textfor,n,textlarge\\
textinconclusive&,textotherwise
endcases$$
The root test is stronger than the ratio test since
$$liminf_ntoinftyleft|fraca_n+1a_nright|le liminf_nto inftysqrt[n]le limsup_ntoinftysqrt[n]le limsup_ntoinftyleft|fraca_n+1a_nright|$$
In the example in the OP, the ratio test in inconclusive since $L=infty$ and $ell=0$. However, the root test reveals
$$limsup_ntoinftysqrt[n]=2^-1/2<1$$
and the series converges.
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
The ratio test can be expressed as follows.
Let $ell=liminf_nto inftyleft|fraca_n+1a_nright|$ and let $L=limsup_nto inftyleft|fraca_n+1a_nright|$. Then, the series $$sum_n=1^infty a_nbegincases
textconverges (absolutey)&, L<1\\
textdiverges &, ell>1\\
textdiverges &, left|fraca_n+1a_nright|ge1,textfor,n,textlarge\\
textinconclusive&,textotherwise
endcases$$
The root test is stronger than the ratio test since
$$liminf_ntoinftyleft|fraca_n+1a_nright|le liminf_nto inftysqrt[n]le limsup_ntoinftysqrt[n]le limsup_ntoinftyleft|fraca_n+1a_nright|$$
In the example in the OP, the ratio test in inconclusive since $L=infty$ and $ell=0$. However, the root test reveals
$$limsup_ntoinftysqrt[n]=2^-1/2<1$$
and the series converges.
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
$endgroup$
add a comment |
$begingroup$
The ratio test can be expressed as follows.
Let $ell=liminf_nto inftyleft|fraca_n+1a_nright|$ and let $L=limsup_nto inftyleft|fraca_n+1a_nright|$. Then, the series $$sum_n=1^infty a_nbegincases
textconverges (absolutey)&, L<1\\
textdiverges &, ell>1\\
textdiverges &, left|fraca_n+1a_nright|ge1,textfor,n,textlarge\\
textinconclusive&,textotherwise
endcases$$
The root test is stronger than the ratio test since
$$liminf_ntoinftyleft|fraca_n+1a_nright|le liminf_nto inftysqrt[n]le limsup_ntoinftysqrt[n]le limsup_ntoinftyleft|fraca_n+1a_nright|$$
In the example in the OP, the ratio test in inconclusive since $L=infty$ and $ell=0$. However, the root test reveals
$$limsup_ntoinftysqrt[n]=2^-1/2<1$$
and the series converges.
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
$endgroup$
The ratio test can be expressed as follows.
Let $ell=liminf_nto inftyleft|fraca_n+1a_nright|$ and let $L=limsup_nto inftyleft|fraca_n+1a_nright|$. Then, the series $$sum_n=1^infty a_nbegincases
textconverges (absolutey)&, L<1\\
textdiverges &, ell>1\\
textdiverges &, left|fraca_n+1a_nright|ge1,textfor,n,textlarge\\
textinconclusive&,textotherwise
endcases$$
The root test is stronger than the ratio test since
$$liminf_ntoinftyleft|fraca_n+1a_nright|le liminf_nto inftysqrt[n]le limsup_ntoinftysqrt[n]le limsup_ntoinftyleft|fraca_n+1a_nright|$$
In the example in the OP, the ratio test in inconclusive since $L=infty$ and $ell=0$. However, the root test reveals
$$limsup_ntoinftysqrt[n]=2^-1/2<1$$
and the series converges.
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
edited Mar 22 at 16:19
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
answered Mar 22 at 15:58
Mark ViolaMark Viola
134k1278177
134k1278177
add a comment |
add a comment |
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$begingroup$
Yes. Look carefully at the proof of the ratio test and you'll see the limit is not required to exist.
$endgroup$
– Umberto P.
Mar 22 at 15:49
$begingroup$
@new2math Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Apr 3 at 3:31