Spivak (sequence)
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here? I just started sequences the other day, so forgive me if this is an easy problem.
calculus
|
show 5 more comments
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here? I just started sequences the other day, so forgive me if this is an easy problem.
calculus
2
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
1
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
1
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
1
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago
|
show 5 more comments
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here? I just started sequences the other day, so forgive me if this is an easy problem.
calculus
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here? I just started sequences the other day, so forgive me if this is an easy problem.
calculus
calculus
edited 1 hour ago
kyle campbell
asked 2 hours ago
kyle campbellkyle campbell
414
414
2
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
1
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
1
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
1
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago
|
show 5 more comments
2
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
1
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
1
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
1
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago
2
2
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
1
1
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
1
1
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
1
1
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago
|
show 5 more comments
2 Answers
2
active
oldest
votes
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
add a comment |
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
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%2f3068257%2fspivak-sequence%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
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
add a comment |
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
add a comment |
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 1 hour ago
Tyler6Tyler6
699212
699212
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
add a comment |
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
1
1
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
Indeed! Thank you for that input.
– kyle campbell
1 hour ago
add a comment |
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
add a comment |
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
add a comment |
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 39 mins ago
MustangMustang
2927
2927
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3068257%2fspivak-sequence%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
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
– Lord Shark the Unknown
2 hours ago
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
1 hour ago
1
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
1 hour ago
1
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
1 hour ago
1
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
1 hour ago