Convergenge or divergence of series with eCalculus(Convergence/Divergence of series)The Series( $sum_1^+ inftyfrac1n! + n$ convergence or divergence?Good source of worked examples for Testing Infinite SeriesConvergence of the series $sum_n=1^inftycot^-1(n)$Proof divergence of a seriesWill this series with radical converge?Divergence/convergence of the seriesDetermine if the infinite series converge or divergeProving convergence or divergence of a series$sum_n=1^infty lnleft(fracnn+1right)$ convergence or divergence
Reverse of diffraction
Who gets an Apparition licence?
Most elegant way to write a one shot IF
Do space suits measure "methane" levels or other biological gases?
What is "oversubscription" in Networking?
What's the easiest way for a whole party to be able to communicate with a creature that doesn't know Common?
Is there a way for presidents to legally extend their terms beyond the maximum of four years?
Should I share with a new service provider a bill from its competitor?
How did researchers use to find articles before the Internet and the computer era?
Needle Hotend for nonplanar printing
Using aluminium busbar/cables in an aircraft instead of copper
One folder two different locations on ubuntu 18.04
Acceleration in Circular motion
Can you sign using a digital signature itself?
Are metaheuristics ever practical for continuous optimization?
Do I have to roll to maintain concentration if a target other than me who is affected by my concentration spell takes damage?
Does “comme on était à New York” mean “since” or “as though”?
Can I ask to speak to my future colleagues before accepting an offer?
Miss Toad and her frogs
Is it bad to describe a character long after their introduction?
Why are 120 V general receptacle circuits limited to 20 A?
How was film developed in the late 1920s?
What does Mildred mean by this line in Three Billboards Outside Ebbing, Missouri?
Can the passive "être + verbe" sometimes mean the past?
Convergenge or divergence of series with e
Calculus(Convergence/Divergence of series)The Series( $sum_1^+ inftyfrac1n! + n$ convergence or divergence?Good source of worked examples for Testing Infinite SeriesConvergence of the series $sum_n=1^inftycot^-1(n)$Proof divergence of a seriesWill this series with radical converge?Divergence/convergence of the seriesDetermine if the infinite series converge or divergeProving convergence or divergence of a series$sum_n=1^infty lnleft(fracnn+1right)$ convergence or divergence
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I can't find the convergence or divergence of the following series by using aspect ratio test or comparison test. The series is:
$$ sum_n=1^infty e^-n^2$$
Thanks.
real-analysis calculus sequences-and-series algebra-precalculus
real-analysis calculus sequences-and-series algebra-precalculus
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
asked 8 hours ago
Un Chico MásUn Chico Más
121 bronze badge
121 bronze badge
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Un Chico Más is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
$endgroup$
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
$endgroup$
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
$endgroup$
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
);
);
Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3272852%2fconvergenge-or-divergence-of-series-with-e%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
$endgroup$
Since$$(forall ninmathbb N):sqrt[n]e^-n^2=e^-nto0,$$the series converges, by the root test.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
edited 8 hours ago
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
195k24 gold badges153 silver badges270 bronze badges
195k24 gold badges153 silver badges270 bronze badges
add a comment |
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
$endgroup$
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
$endgroup$
add a comment |
$begingroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
$endgroup$
Since $e^-n^2leq e^-n$, the series converges by comparison with the geometric series $sum_n=1^infty e^-n$
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
answered 8 hours ago
saulspatzsaulspatz
20.4k4 gold badges16 silver badges36 bronze badges
20.4k4 gold badges16 silver badges36 bronze badges
add a comment |
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Also by ratio test:
$$
left|fraca_n+1a_nright| = frac1e^2n+1 to 0 text as $n to infty$
$$
Also by comparison test:
$$
e^n^2 > n^2 implies frac1e^n^2 < frac1n^2
$$
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
answered 8 hours ago
Ruben du BurckRuben du Burck
5962 silver badges11 bronze badges
5962 silver badges11 bronze badges
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ruben du Burck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
$endgroup$
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
$endgroup$
add a comment |
$begingroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
$endgroup$
The series converges rather violently.
Since
$n^2 ge n$,
$e^-n^2 le e^-n$,
and the sum of that converges
so your series also converges.
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
answered 8 hours ago
marty cohenmarty cohen
77.7k5 gold badges49 silver badges133 bronze badges
77.7k5 gold badges49 silver badges133 bronze badges
add a comment |
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
$endgroup$
add a comment |
$begingroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
$endgroup$
It converges also via the ratio test :
$frace^-(n+1)^2e^-n^2 = e^-2n - 1 rightarrow 0$, which is a limit of absolute value $< 1$.
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
answered 8 hours ago
DLeMeurDLeMeur
4599 bronze badges
4599 bronze badges
add a comment |
add a comment |
Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.
Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.
Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.
Un Chico Más is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3272852%2fconvergenge-or-divergence-of-series-with-e%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
