Fdtxjr

Why can't we play rap on piano?

Can a Cauchy sequence converge for one metric while not converging for another?

What is the word for reserving something for yourself before others do?

dbcc cleantable batch size explanation

Languages that we cannot (dis)prove to be Context-Free

Has there ever been an airliner design involving reducing generator load by installing solar panels?

Is it legal for company to use my work email to pretend I still work there?

meaning of に in 本当に?

Why does Kotter return in Welcome Back Kotter?

What defenses are there against being summoned by the Gate spell?

Convert two switches to a dual stack, and add outlet - possible here?

Could an aircraft fly or hover using only jets of compressed air?

How do I deal with an unproductive colleague in a small company?

What does it mean to describe someone as a butt steak?

How does one intimidate enemies without having the capacity for violence?

Are the number of citations and number of published articles the most important criteria for a tenure promotion?

High voltage LED indicator 40-1000 VDC without additional power supply

How can I prevent hyper evolved versions of regular creatures from wiping out their cousins?

Was any UN Security Council vote triple-vetoed?

Revoked SSL certificate

A case of the sniffles

What doth I be?

RSA: Danger of using p to create q

Filter any system log file by date or date range

Help with proof of Root Test.

Proof of convergenceConvergence of $sumlimits_n=1^inftyfrac1+(-1)^nn$Show by comparison that $sumlimits_n=1^infty sin(frac1n)$ diverges?relations between the root test and the ratio testHelp with proving that this series divergesShow that $sum a_n$ diverges if $sum log (tfrac11-a_n)$ divergesconvergent and divergent series indeterminate by ratio and root testShow that a Series DivergesProof of Root Test?Intuition for Root Test.

1

$begingroup$

I'm trying to prove the case,

Let $limlimits_n to ∞ sqrt[n] = L>1$ then the series $displaystylesum a_n$ is divergent.

My Attempt :

Let $r$ be a number such that $L>r>1$ and as $|a_n|^1/n to L$ as $n to ∞$ then there exists an $m in mathbbN$ such that for all $n≥m, |a_n|^1/n>r>1 Rightarrow |a_n|> r^n$ and $ displaystylesum r^n$ diverges so by comparison test $displaystylesum |a_n|$ diverges.

Now how do I prove that $displaystylesum a_n $ diverges?

Also if there are any mistakes in my prove, do point out. Thanks.

sequences-and-series

share|cite|improve this question

asked Mar 21 at 17:44

William William

1,230414

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг Correct me if I'm wrong, the contrapositive will be "If $L>1$ then $sum a_n$ is NOT convergent. But it still doesn't prove that it is divergent as $sum a_n$ is not necessarily a series of positive terms right? ...
  $endgroup$
  – William
  Mar 21 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My fault : I thought divergent is the same as "not convergent", I think that is a mistake. If divergent means going to plus/minus infinity then we will need a different argument.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 18:09
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг No worries, that is exactly what's been bothering me. I have two Real Analysis books. In one they have mentioned Root Test with series of positive terms which is straightforward as you don't have absolute values. But in the other book and also on the internet, I found this, and I've been racking my brains ever since only to make sense out of this.
  $endgroup$
  – William
  Mar 21 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with you. However, if you are for example taking a course, then your teacher should have the final say in this matter. This business has been bothering me a bit : when it comes up in a conversation I just clarify the meaning with whoever brought it up and move on, relieved that my teacher stuck with divergent meaning "not convergent" in my days.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг No unfortunately I don't have teachers at all, I'm just self studying from different books, it's hard to understand everything on your own, but I'm glad I'm born in the age of Internet.
  $endgroup$
  – William
  Mar 21 at 18:27
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I'm trying to prove the case,

Let $limlimits_n to ∞ sqrt[n] = L>1$ then the series $displaystylesum a_n$ is divergent.

My Attempt :

Let $r$ be a number such that $L>r>1$ and as $|a_n|^1/n to L$ as $n to ∞$ then there exists an $m in mathbbN$ such that for all $n≥m, |a_n|^1/n>r>1 Rightarrow |a_n|> r^n$ and $ displaystylesum r^n$ diverges so by comparison test $displaystylesum |a_n|$ diverges.

Now how do I prove that $displaystylesum a_n $ diverges?

Also if there are any mistakes in my prove, do point out. Thanks.

sequences-and-series

share|cite|improve this question

asked Mar 21 at 17:44

William William

1,230414

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг Correct me if I'm wrong, the contrapositive will be "If $L>1$ then $sum a_n$ is NOT convergent. But it still doesn't prove that it is divergent as $sum a_n$ is not necessarily a series of positive terms right? ...
  $endgroup$
  – William
  Mar 21 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My fault : I thought divergent is the same as "not convergent", I think that is a mistake. If divergent means going to plus/minus infinity then we will need a different argument.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 18:09
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг No worries, that is exactly what's been bothering me. I have two Real Analysis books. In one they have mentioned Root Test with series of positive terms which is straightforward as you don't have absolute values. But in the other book and also on the internet, I found this, and I've been racking my brains ever since only to make sense out of this.
  $endgroup$
  – William
  Mar 21 at 18:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I agree with you. However, if you are for example taking a course, then your teacher should have the final say in this matter. This business has been bothering me a bit : when it comes up in a conversation I just clarify the meaning with whoever brought it up and move on, relieved that my teacher stuck with divergent meaning "not convergent" in my days.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 21 at 18:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @астонвіллаолофмэллбэрг No unfortunately I don't have teachers at all, I'm just self studying from different books, it's hard to understand everything on your own, but I'm glad I'm born in the age of Internet.
  $endgroup$
  – William
  Mar 21 at 18:27
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I'm trying to prove the case,

Let $limlimits_n to ∞ sqrt[n] = L>1$ then the series $displaystylesum a_n$ is divergent.

My Attempt :

Let $r$ be a number such that $L>r>1$ and as $|a_n|^1/n to L$ as $n to ∞$ then there exists an $m in mathbbN$ such that for all $n≥m, |a_n|^1/n>r>1 Rightarrow |a_n|> r^n$ and $ displaystylesum r^n$ diverges so by comparison test $displaystylesum |a_n|$ diverges.

Now how do I prove that $displaystylesum a_n $ diverges?

Also if there are any mistakes in my prove, do point out. Thanks.

sequences-and-series

share|cite|improve this question

asked Mar 21 at 17:44

William William

1,230414

$endgroup$

share|cite|improve this question

asked Mar 21 at 17:44

William William

1,230414

share|cite|improve this question

asked Mar 21 at 17:44

William William

1,230414

asked Mar 21 at 17:44

William William

1,230414

asked Mar 21 at 17:44

William William

1,230414

2 Answers
2

active

oldest

votes

0

$begingroup$

Suppose for the sake of contradiction that $sum a_n$ is convergent. Then, the limit $lim_n rightarrow infty a_n = 0$.

Now, since $|a_n| > 0$ for all $n$, then the operation $(a_n)^1/n$ is continuous, so you can pass the limit through the $n$th root. Thus $lim_n rightarrow infty |a_n|^1/n = |lim_n rightarrow infty a_n |^1/n = 0^1/n = 0$. Then you've reached a contradiction.

share|cite|improve this answer

answered Mar 21 at 17:55

kkckkc

17011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the conclusion would be that the series is NOT convergent. We still haven't proved that it is divergent. Have we? Because $sum a_n $ is not necessarily a series of positive terms so... Yea
  $endgroup$
  – William
  Mar 21 at 18:10
  

  
  
  
  

add a comment | 

0

$begingroup$

$sum_n=1^infty a_n > sum_n=1^infty sqrt[n]a_n$

for a given $n>k$ and $epsilon >0$ arbitrary small

$sum_n=k^infty sqrt[n]a_n >sum_n=k^infty L - epsilon geq sum_n=k^infty 1 = infty $

share|cite|improve this answer

answered Mar 21 at 18:07

user1828958user1828958

667

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3157124%2fhelp-with-proof-of-root-test%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Suppose for the sake of contradiction that $sum a_n$ is convergent. Then, the limit $lim_n rightarrow infty a_n = 0$.

Now, since $|a_n| > 0$ for all $n$, then the operation $(a_n)^1/n$ is continuous, so you can pass the limit through the $n$th root. Thus $lim_n rightarrow infty |a_n|^1/n = |lim_n rightarrow infty a_n |^1/n = 0^1/n = 0$. Then you've reached a contradiction.

share|cite|improve this answer

answered Mar 21 at 17:55

kkckkc

17011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the conclusion would be that the series is NOT convergent. We still haven't proved that it is divergent. Have we? Because $sum a_n $ is not necessarily a series of positive terms so... Yea
  $endgroup$
  – William
  Mar 21 at 18:10
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose for the sake of contradiction that $sum a_n$ is convergent. Then, the limit $lim_n rightarrow infty a_n = 0$.

Now, since $|a_n| > 0$ for all $n$, then the operation $(a_n)^1/n$ is continuous, so you can pass the limit through the $n$th root. Thus $lim_n rightarrow infty |a_n|^1/n = |lim_n rightarrow infty a_n |^1/n = 0^1/n = 0$. Then you've reached a contradiction.

share|cite|improve this answer

answered Mar 21 at 17:55

kkckkc

17011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So the conclusion would be that the series is NOT convergent. We still haven't proved that it is divergent. Have we? Because $sum a_n $ is not necessarily a series of positive terms so... Yea
  $endgroup$
  – William
  Mar 21 at 18:10
  

  
  
  
  

add a comment | 

$begingroup$

Suppose for the sake of contradiction that $sum a_n$ is convergent. Then, the limit $lim_n rightarrow infty a_n = 0$.

Now, since $|a_n| > 0$ for all $n$, then the operation $(a_n)^1/n$ is continuous, so you can pass the limit through the $n$th root. Thus $lim_n rightarrow infty |a_n|^1/n = |lim_n rightarrow infty a_n |^1/n = 0^1/n = 0$. Then you've reached a contradiction.

share|cite|improve this answer

answered Mar 21 at 17:55

kkckkc

17011

$endgroup$

share|cite|improve this answer

answered Mar 21 at 17:55

kkckkc

17011

answered Mar 21 at 17:55

kkckkc

17011

answered Mar 21 at 17:55

kkckkc

17011

0

$begingroup$

$sum_n=1^infty a_n > sum_n=1^infty sqrt[n]a_n$

for a given $n>k$ and $epsilon >0$ arbitrary small

$sum_n=k^infty sqrt[n]a_n >sum_n=k^infty L - epsilon geq sum_n=k^infty 1 = infty $

share|cite|improve this answer

answered Mar 21 at 18:07

user1828958user1828958

667

$endgroup$

add a comment | 

0

$begingroup$

$sum_n=1^infty a_n > sum_n=1^infty sqrt[n]a_n$

for a given $n>k$ and $epsilon >0$ arbitrary small

$sum_n=k^infty sqrt[n]a_n >sum_n=k^infty L - epsilon geq sum_n=k^infty 1 = infty $

share|cite|improve this answer

answered Mar 21 at 18:07

user1828958user1828958

667

$endgroup$

add a comment | 

$begingroup$

$sum_n=1^infty a_n > sum_n=1^infty sqrt[n]a_n$

for a given $n>k$ and $epsilon >0$ arbitrary small

$sum_n=k^infty sqrt[n]a_n >sum_n=k^infty L - epsilon geq sum_n=k^infty 1 = infty $

share|cite|improve this answer

answered Mar 21 at 18:07

user1828958user1828958

667

$endgroup$

share|cite|improve this answer

answered Mar 21 at 18:07

user1828958user1828958

667

answered Mar 21 at 18:07

user1828958user1828958

667

answered Mar 21 at 18:07

user1828958user1828958

667