When to use the root test. Is this not a good situation to use it? The 2019 Stack Overflow Developer Survey Results Are InWhich test would be appropriate to use on this series to show convergence/divergence?Integral test vs root test vs ratio testHow to show convergence or divergence of a series when the ratio test is inconclusive?Root test with nested power function?Confused about using alternating test, ratio test, and root test (please help).Radius and interval of convergence of $sum_n=1^infty(-1)^nfracx^2n(2n)!$ by root and ratio test are different?How would I use root/ratio test on $sum_n=1^inftyleft(fracnn+1right)^n^2$?How would I know when to use what test for convergence?convergence of a sum fails with root testIntuition for Root Test.
What is the meaning of Triage in Cybersec world?
Is it correct to say the Neural Networks are an alternative way of performing Maximum Likelihood Estimation? if not, why?
Does HR tell a hiring manager about salary negotiations?
What is the light source in the black hole images?
Pokemon Turn Based battle (Python)
Loose spokes after only a few rides
Old scifi movie from the 50s or 60s with men in solid red uniforms who interrogate a spy from the past
Why does the nucleus not repel itself?
Straighten subgroup lattice
Getting crown tickets for Statue of Liberty
Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?
Merge two greps into single one
How to obtain a position of last non-zero element
What is this sharp, curved notch on my knife for?
Compute the product of 3 dictionaries and concatenate keys and values
Is it safe to harvest rainwater that fell on solar panels?
The phrase "to the numbers born"?
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
What's the name of these plastic connectors
Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?
What do hard-Brexiteers want with respect to the Irish border?
A word that means fill it to the required quantity
Deal with toxic manager when you can't quit
How do PCB vias affect signal quality?
When to use the root test. Is this not a good situation to use it?
The 2019 Stack Overflow Developer Survey Results Are InWhich test would be appropriate to use on this series to show convergence/divergence?Integral test vs root test vs ratio testHow to show convergence or divergence of a series when the ratio test is inconclusive?Root test with nested power function?Confused about using alternating test, ratio test, and root test (please help).Radius and interval of convergence of $sum_n=1^infty(-1)^nfracx^2n(2n)!$ by root and ratio test are different?How would I use root/ratio test on $sum_n=1^inftyleft(fracnn+1right)^n^2$?How would I know when to use what test for convergence?convergence of a sum fails with root testIntuition for Root Test.
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked yesterday
Jwan622Jwan622
2,38711632
$endgroup$
add a comment |
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked yesterday
Jwan622Jwan622
2,38711632
$endgroup$
add a comment |
$begingroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
asked yesterday
Jwan622Jwan622
2,38711632
$endgroup$
I'm having trouble seeing when to use the root test. nth powers occur, but I think the ratio test is easier:
Here is the problem:
$$sum_n=1^infty fracx^nn^44^n$$
So the ratio test seems to work here, but can't the root test be used to? The problem is that the $n^4$ doesnt play well with the root test right?
Here is the beginning of my solution with the ratio test:
$$biggr lbrack fraca_n+1a_n biggr rbrack = biggr lbrack fracx^n+1(n+1)^4 * 4^n+1 * fracn^4*4^nx^n biggr rbrack = biggr lbrack fracx*n^4(n+1)^4 * 4 biggr rbrack = fracx4$$
So I don't think the explanation for when to use the root test is totally right right? I can't really use it here because the $n^4$ causes some problems with the root test right?
sequences-and-series
sequences-and-series
asked yesterday
Jwan622Jwan622
2,38711632
asked yesterday
Jwan622Jwan622
2,38711632
asked yesterday
Jwan622Jwan622
2,38711632
asked yesterday
Jwan622Jwan622
2,38711632
asked yesterday
Jwan622Jwan622
2,38711632
2,38711632
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered yesterday
marty cohenmarty cohen
75.2k549130
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered yesterday
MelodyMelody
1,09412
$endgroup$
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181802%2fwhen-to-use-the-root-test-is-this-not-a-good-situation-to-use-it%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered yesterday
marty cohenmarty cohen
75.2k549130
$endgroup$
add a comment |
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered yesterday
marty cohenmarty cohen
75.2k549130
$endgroup$
add a comment |
$begingroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered yesterday
marty cohenmarty cohen
75.2k549130
$endgroup$
When doing a root test,
powers of $n$ can be ignored
because,
for any fixed $k$,
$lim_n to infty (n^k)^1/n
=1
$.
This is because
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
$
and
$lim_n to infty fracln(n)n
=0$.
An easy,
but nonelementary proof of this is this:
$beginarray\
ln(n)
&=int_1^n dfracdtt\
&<int_1^n dfracdtt^1/2\
&=2t^1/2|_1^n\
< 2sqrtn\
textso\
dfracln(n)n
&<dfrac2sqrtn\
endarray
$
Therefore
$ (n^k)^1/n
=n^k/n
=e^k ln(n)/n
lt e^2k/sqrtn
to 1
$.
answered yesterday
marty cohenmarty cohen
75.2k549130
answered yesterday
marty cohenmarty cohen
75.2k549130
answered yesterday
marty cohenmarty cohen
75.2k549130
answered yesterday
marty cohenmarty cohen
75.2k549130
75.2k549130
add a comment |
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered yesterday
MelodyMelody
1,09412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered yesterday
MelodyMelody
1,09412
$endgroup$
add a comment |
$begingroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered yesterday
MelodyMelody
1,09412
$endgroup$
It doesn't cause any problems, because $lim_ntoinftysqrt[n]n^4=1.$ Actually, the root test is stronger than the ratio test. Sometimes the root test limit exists, but the ratio test limit does not. However, if they both exist, then they are equal. Which is why if one limit is $1$ you shouldn't try the other, even though the root test is stronger.
answered yesterday
MelodyMelody
1,09412
answered yesterday
MelodyMelody
1,09412
answered yesterday
MelodyMelody
1,09412
answered yesterday
MelodyMelody
1,09412
1,09412
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181802%2fwhen-to-use-the-root-test-is-this-not-a-good-situation-to-use-it%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
