Clarification on step in proof for Ratio test
$begingroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
$endgroup$
add a comment |
$begingroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
$endgroup$
add a comment |
$begingroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
$endgroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
sequences-and-series convergence limsup-and-liminf
asked Dec 20 '18 at 11:06
CruZCruZ
527
527
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
$endgroup$
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
$endgroup$
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%2f3047425%2fclarification-on-step-in-proof-for-ratio-test%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
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
$endgroup$
add a comment |
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
$endgroup$
add a comment |
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
$endgroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
21718
add a comment |
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
$endgroup$
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
$endgroup$
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
$endgroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,745213
1,745213
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%2f3047425%2fclarification-on-step-in-proof-for-ratio-test%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