Prove that the limit exists or does not existHow to Prove the Root Test For SeriesProving that $sum fracn^n+1/n(n+1/n)^n$ divergesShow that there exists a real function $f$ such that $xto a, |f(x)|to |L|$ but the limit of $f(x)$ does not exist.Convergence/Divergence of the series $sumlimits_n=1^inftytan(1/n)$Prove the series $sum_n=1^∞ (-1)^n(n)/(n+2)$ divergesL'Hopital rule to show limit does not existDoes the following limit exist? $lim_rto1sum_n=0^infty a_nr^n$Limit of the composistion of two functions when the limit of $f$ does not exist?Does the series $sum_n=1^infty fracnsqrt[3]8n^5-1$ Converge?If $lim textsup_ntoinftyX_n≠lim textinf_ntoinftyX_n$, can we say “limit doesn't exist”?
Pressure inside an infinite ocean?
Hyperlink on red background
Set collection doesn't always enforce uniqueness with the Date datatype? Does the following example seem correct?
Should I replace my bicycle tires if they have not been inflated in multiple years
Would Hubble Space Telescope improve black hole image observed by EHT if it joined array of telesopes?
How do LIGO and VIRGO know that a gravitational wave has its origin in a neutron star or a black hole?
Shantae Dance Matching
Verb "geeitet" in an old scientific text
Independent, post-Brexit Scotland - would there be a hard border with England?
How long would it take for people to notice a mass disappearance?
Why is the relative clause in the following sentence not directly after the noun and why is the verb not in the end of the sentence?
I'm in your subnets, golfing your code
Would the Disguise Self spell be able to reveal hidden birthmarks/tattoos (of the person they're disguised as) to a character?
Point of the the Dothraki's attack in GoT S8E3?
Can Infinity Stones be retrieved more than once?
Expressing 'our' for objects belonging to our apartment
Double or Take game
Multi-channel audio upsampling interpolation
Why Isn’t SQL More Refactorable?
How important is people skills in academic career and applications?
Building a list of products from the elements in another list
What matters more when it comes to book covers? Is it ‘professional quality’ or relevancy?
Why was the battle set up *outside* Winterfell?
Have I damaged my car by attempting to reverse with hand/park brake up?
Prove that the limit exists or does not exist
How to Prove the Root Test For SeriesProving that $sum fracn^n+1/n(n+1/n)^n$ divergesShow that there exists a real function $f$ such that $xto a, |f(x)|to |L|$ but the limit of $f(x)$ does not exist.Convergence/Divergence of the series $sumlimits_n=1^inftytan(1/n)$Prove the series $sum_n=1^∞ (-1)^n(n)/(n+2)$ divergesL'Hopital rule to show limit does not existDoes the following limit exist? $lim_rto1sum_n=0^infty a_nr^n$Limit of the composistion of two functions when the limit of $f$ does not exist?Does the series $sum_n=1^infty fracnsqrt[3]8n^5-1$ Converge?If $lim textsup_ntoinftyX_n≠lim textinf_ntoinftyX_n$, can we say “limit doesn't exist”?
$begingroup$
Prove that the limit exists or does not exist:
$$ lim_Nto infty sum_n=1^N frac1phi(n), $$
where $phi(n)$ is the Euler Totient function.
The ratio test was inconclusive.
I'm fairly sure the p-series test says this series diverges because $p=1$ but then again in this case I'm not sure how to deal with a function in the place where $n$ normally is.
calculus sequences-and-series limits limits-without-lhopital totient-function
asked 2 hours ago
UltradarkUltradark
3891518
$endgroup$
add a comment |
$begingroup$
Prove that the limit exists or does not exist:
$$ lim_Nto infty sum_n=1^N frac1phi(n), $$
where $phi(n)$ is the Euler Totient function.
The ratio test was inconclusive.
I'm fairly sure the p-series test says this series diverges because $p=1$ but then again in this case I'm not sure how to deal with a function in the place where $n$ normally is.
calculus sequences-and-series limits limits-without-lhopital totient-function
asked 2 hours ago
UltradarkUltradark
3891518
$endgroup$
2
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago
add a comment |
$begingroup$
Prove that the limit exists or does not exist:
$$ lim_Nto infty sum_n=1^N frac1phi(n), $$
where $phi(n)$ is the Euler Totient function.
The ratio test was inconclusive.
I'm fairly sure the p-series test says this series diverges because $p=1$ but then again in this case I'm not sure how to deal with a function in the place where $n$ normally is.
calculus sequences-and-series limits limits-without-lhopital totient-function
asked 2 hours ago
UltradarkUltradark
3891518
$endgroup$
Prove that the limit exists or does not exist:
$$ lim_Nto infty sum_n=1^N frac1phi(n), $$
where $phi(n)$ is the Euler Totient function.
The ratio test was inconclusive.
I'm fairly sure the p-series test says this series diverges because $p=1$ but then again in this case I'm not sure how to deal with a function in the place where $n$ normally is.
calculus sequences-and-series limits limits-without-lhopital totient-function
calculus sequences-and-series limits limits-without-lhopital totient-function
asked 2 hours ago
UltradarkUltradark
3891518
asked 2 hours ago
UltradarkUltradark
3891518
asked 2 hours ago
UltradarkUltradark
3891518
asked 2 hours ago
UltradarkUltradark
3891518
asked 2 hours ago
UltradarkUltradark
3891518
3891518
2
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago
add a comment |
2
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago
2
2
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $varphi(n)leq n$ it follows that $$frac1varphi(n)geq frac1n$$ and hence $sum frac1varphi(n)=infty$.
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
$endgroup$
add a comment |
$begingroup$
A way of proof is the following:
$dfrac1p-1gt dfrac1p$ and $phi(p)=p-1$ then $$sum_n=1^N dfrac1phi(n)gtsum_ptext prime^N dfrac1phi(p)gtsum_ptext prime^N dfrac1pto infty$$ (because of the Euler's proof of the infinity of primes).
answered 1 hour ago
PiquitoPiquito
18.1k31540
$endgroup$
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
add a comment |
Your Answer
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%2f3210362%2fprove-that-the-limit-exists-or-does-not-exist%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
$begingroup$
Since $varphi(n)leq n$ it follows that $$frac1varphi(n)geq frac1n$$ and hence $sum frac1varphi(n)=infty$.
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
$endgroup$
add a comment |
$begingroup$
Since $varphi(n)leq n$ it follows that $$frac1varphi(n)geq frac1n$$ and hence $sum frac1varphi(n)=infty$.
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
$endgroup$
add a comment |
$begingroup$
Since $varphi(n)leq n$ it follows that $$frac1varphi(n)geq frac1n$$ and hence $sum frac1varphi(n)=infty$.
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
$endgroup$
Since $varphi(n)leq n$ it follows that $$frac1varphi(n)geq frac1n$$ and hence $sum frac1varphi(n)=infty$.
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
answered 2 hours ago
Foobaz JohnFoobaz John
23.6k41553
23.6k41553
add a comment |
add a comment |
$begingroup$
A way of proof is the following:
$dfrac1p-1gt dfrac1p$ and $phi(p)=p-1$ then $$sum_n=1^N dfrac1phi(n)gtsum_ptext prime^N dfrac1phi(p)gtsum_ptext prime^N dfrac1pto infty$$ (because of the Euler's proof of the infinity of primes).
answered 1 hour ago
PiquitoPiquito
18.1k31540
$endgroup$
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
add a comment |
$begingroup$
A way of proof is the following:
$dfrac1p-1gt dfrac1p$ and $phi(p)=p-1$ then $$sum_n=1^N dfrac1phi(n)gtsum_ptext prime^N dfrac1phi(p)gtsum_ptext prime^N dfrac1pto infty$$ (because of the Euler's proof of the infinity of primes).
answered 1 hour ago
PiquitoPiquito
18.1k31540
$endgroup$
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
add a comment |
$begingroup$
A way of proof is the following:
$dfrac1p-1gt dfrac1p$ and $phi(p)=p-1$ then $$sum_n=1^N dfrac1phi(n)gtsum_ptext prime^N dfrac1phi(p)gtsum_ptext prime^N dfrac1pto infty$$ (because of the Euler's proof of the infinity of primes).
answered 1 hour ago
PiquitoPiquito
18.1k31540
$endgroup$
A way of proof is the following:
$dfrac1p-1gt dfrac1p$ and $phi(p)=p-1$ then $$sum_n=1^N dfrac1phi(n)gtsum_ptext prime^N dfrac1phi(p)gtsum_ptext prime^N dfrac1pto infty$$ (because of the Euler's proof of the infinity of primes).
answered 1 hour ago
PiquitoPiquito
18.1k31540
answered 1 hour ago
PiquitoPiquito
18.1k31540
answered 1 hour ago
PiquitoPiquito
18.1k31540
answered 1 hour ago
PiquitoPiquito
18.1k31540
18.1k31540
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
add a comment |
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Infinity of primes is not enough to conclude that the last sum diverges. There are infinitely many numbers of the form $n^2$ or $2^n$ but those sums converge.... You probably need the Prime number theorem for the last step, which is pretty heavy.
$endgroup$
– N. S.
1 hour ago
1
1
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour ago
$begingroup$
He uses a stronger version : it's not just the infinity of primes itself, but Euler's proof of that result, which actually shows this sum diverges (so there cannot be a finite number of terms). The proof is not the full on Prime Number Theorem (you need Euler's product formula, and the fact that $lim_x to 1^+ zeta(x) = +infty$).
$endgroup$
– Joel Cohen
1 hour 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%2f3210362%2fprove-that-the-limit-exists-or-does-not-exist%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
2
$begingroup$
$phi(n) leq n$ might help.
$endgroup$
– Jo'
2 hours ago