Show that $limsup_{k to infty} 2^{-k} N_k = 0$ where $N_k$ is the number of $a_n geq 2^{-k}$.
$begingroup$
Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.
Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.
real-analysis analysis convergence summation limsup-and-liminf
asked Dec 10 '18 at 21:46
ReaveredReavered
161
$endgroup$
add a comment |
$begingroup$
Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.
Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.
real-analysis analysis convergence summation limsup-and-liminf
asked Dec 10 '18 at 21:46
ReaveredReavered
161
$endgroup$
add a comment |
$begingroup$
Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.
Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.
real-analysis analysis convergence summation limsup-and-liminf
asked Dec 10 '18 at 21:46
ReaveredReavered
161
$endgroup$
Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.
Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.
real-analysis analysis convergence summation limsup-and-liminf
real-analysis analysis convergence summation limsup-and-liminf
asked Dec 10 '18 at 21:46
ReaveredReavered
161
asked Dec 10 '18 at 21:46
ReaveredReavered
161
asked Dec 10 '18 at 21:46
ReaveredReavered
161
asked Dec 10 '18 at 21:46
ReaveredReavered
161
asked Dec 10 '18 at 21:46
ReaveredReavered
161
161
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
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%2f3034530%2fshow-that-limsup-k-to-infty-2-k-n-k-0-where-n-k-is-the-number-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
add a comment |
$begingroup$
Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
add a comment |
$begingroup$
Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Dec 10 '18 at 23:35
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
57.6k42160
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%2f3034530%2fshow-that-limsup-k-to-infty-2-k-n-k-0-where-n-k-is-the-number-of%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
