Does this power sequence converge or diverge? If it converges, what is the limit?Why does this pattern fail (sometimes) for the continued fraction convergents of $sqrt2$?Does this sequence always give a square number?Does the sequence $A_n = fracsin (n)sqrtn$ converge to $0$?Do this series converge or diverge?Does this sequence converge $a_n=frac 3^n+25^n $Given: $sum n a_n$ is convergent. To prove: The sequence $a_n$ convergesHow do you find the value of $sum_r=1^infty frac6^r(3^r-2^r)(3^r+1 - 2^r+1) $?Confirming that the sequence $a_n = fracsqrtncosnsqrtn^3-1$ convergesDoes this non-monotonic sequence converge?Finding the limit of a sequence involving n-th roots.
Why escape if the_content isnt?
Is there a problem with hiding "forgot password" until it's needed?
Applicability of Single Responsibility Principle
What is paid subscription needed for in Mortal Kombat 11?
Closest Prime Number
A particular customize with green line and letters for subfloat
Is this version of a gravity generator feasible?
Escape a backup date in a file name
Is expanding the research of a group into machine learning as a PhD student risky?
How to Reset Passwords on Multiple Websites Easily?
Is there a good way to store credentials outside of a password manager?
Why are there no referendums in the US?
Lay out the Carpet
Did Dumbledore lie to Harry about how long he had James Potter's invisibility cloak when he was examining it? If so, why?
How does buying out courses with grant money work?
Why not increase contact surface when reentering the atmosphere?
Balance Issues for a Custom Sorcerer Variant
Is `x >> pure y` equivalent to `liftM (const y) x`
Was Spock the First Vulcan in Starfleet?
Valid Badminton Score?
Term for the "extreme-extension" version of a straw man fallacy?
What can we do to stop prior company from asking us questions?
How does Loki do this?
How do I find the solutions of the following equation?
Does this power sequence converge or diverge? If it converges, what is the limit?
Why does this pattern fail (sometimes) for the continued fraction convergents of $sqrt2$?Does this sequence always give a square number?Does the sequence $A_n = fracsin (n)sqrtn$ converge to $0$?Do this series converge or diverge?Does this sequence converge $a_n=frac 3^n+25^n $Given: $sum n a_n$ is convergent. To prove: The sequence $a_n$ convergesHow do you find the value of $sum_r=1^infty frac6^r(3^r-2^r)(3^r+1 - 2^r+1) $?Confirming that the sequence $a_n = fracsqrtncosnsqrtn^3-1$ convergesDoes this non-monotonic sequence converge?Finding the limit of a sequence involving n-th roots.
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 6 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
add a comment |
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 6 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago
add a comment |
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 6 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
sequences-and-series
asked 6 hours ago
Jwan622Jwan622
2,30211632
asked 6 hours ago
Jwan622Jwan622
2,30211632
edited 5 hours ago
Jwan622
asked 6 hours ago
Jwan622Jwan622
2,30211632
asked 6 hours ago
Jwan622Jwan622
2,30211632
asked 6 hours ago
Jwan622Jwan622
2,30211632
2,30211632
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago
add a comment |
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
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%2f3164743%2fdoes-this-power-sequence-converge-or-diverge-if-it-converges-what-is-the-limit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
add a comment |
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
add a comment |
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
24.1k2254
add a comment |
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
edited 5 hours ago
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
78.1k42867
add a comment |
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
111
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
add a comment |
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
5 hours ago
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%2f3164743%2fdoes-this-power-sequence-converge-or-diverge-if-it-converges-what-is-the-limit%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
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
5 hours ago