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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?

1

$begingroup$

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

share|cite|improve this question

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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 22 at 16:25

Micah

30.3k1364106

asked Mar 22 at 15:42

New2MathNew2Math

16715

$endgroup$

share|cite|improve this question

edited Mar 22 at 16:25

Micah

30.3k1364106

asked Mar 22 at 15:42

New2MathNew2Math

16715

share|cite|improve this question

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

asked Mar 22 at 15:42

New2MathNew2Math

16715

asked Mar 22 at 15:42

New2MathNew2Math

16715

2 Answers
2

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 22 at 15:49

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 15:58

Mark ViolaMark Viola

134k1278177

$endgroup$

add a comment | 

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0

$begingroup$

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.

share|cite|improve this answer

answered Mar 22 at 15:49

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 22 at 15:49

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 15:58

Mark ViolaMark Viola

134k1278177

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

edited Mar 22 at 16:19

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.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 15:58

Mark ViolaMark Viola

134k1278177

$endgroup$

share|cite|improve this answer

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