Determine whether the sequence $sinfrac{npi}{2}$ converges and prove your answer using the epsilon method..
up vote
1
down vote
favorite
I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.
sequences-and-series limits epsilon-delta
edited Nov 29 at 4:24
Key Flex
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
add a comment |
up vote
1
down vote
favorite
I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.
sequences-and-series limits epsilon-delta
edited Nov 29 at 4:24
Key Flex
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
1
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.
sequences-and-series limits epsilon-delta
edited Nov 29 at 4:24
Key Flex
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.
sequences-and-series limits epsilon-delta
sequences-and-series limits epsilon-delta
edited Nov 29 at 4:24
Key Flex
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
edited Nov 29 at 4:24
Key Flex
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
edited Nov 29 at 4:24
Key Flex
7,44941232
edited Nov 29 at 4:24
Key Flex
7,44941232
edited Nov 29 at 4:24
Key Flex
7,44941232
7,44941232
asked Nov 29 at 3:40
NeedHelp
161
asked Nov 29 at 3:40
NeedHelp
161
asked Nov 29 at 3:40
NeedHelp
161
161
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
1
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47
add a comment |
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
1
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
1
1
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47
add a comment |
4 Answers
4
active
oldest
votes
up vote
2
down vote
HINT:
Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.
What can you conclude now?
answered Nov 29 at 3:52
Mark Viola
130k1273170
add a comment |
up vote
1
down vote
It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
answered Nov 29 at 3:44
Eevee Trainer
3,370225
add a comment |
up vote
1
down vote
If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.
Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$
$sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.
$(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$
What is the length of this interval?
answered Nov 29 at 4:17
Shubham Johri
3,084413
add a comment |
up vote
0
down vote
Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
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%2f3018138%2fdetermine-whether-the-sequence-sin-fracn-pi2-converges-and-prove-your-ans%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT:
Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.
What can you conclude now?
answered Nov 29 at 3:52
Mark Viola
130k1273170
add a comment |
up vote
2
down vote
HINT:
Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.
What can you conclude now?
answered Nov 29 at 3:52
Mark Viola
130k1273170
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT:
Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.
What can you conclude now?
answered Nov 29 at 3:52
Mark Viola
130k1273170
HINT:
Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.
What can you conclude now?
answered Nov 29 at 3:52
Mark Viola
130k1273170
answered Nov 29 at 3:52
Mark Viola
130k1273170
answered Nov 29 at 3:52
Mark Viola
130k1273170
answered Nov 29 at 3:52
Mark Viola
130k1273170
130k1273170
add a comment |
add a comment |
up vote
1
down vote
It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
answered Nov 29 at 3:44
Eevee Trainer
3,370225
add a comment |
up vote
1
down vote
It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
answered Nov 29 at 3:44
Eevee Trainer
3,370225
add a comment |
up vote
1
down vote
up vote
1
down vote
It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
answered Nov 29 at 3:44
Eevee Trainer
3,370225
It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
answered Nov 29 at 3:44
Eevee Trainer
3,370225
answered Nov 29 at 3:44
Eevee Trainer
3,370225
answered Nov 29 at 3:44
Eevee Trainer
3,370225
answered Nov 29 at 3:44
Eevee Trainer
3,370225
3,370225
add a comment |
add a comment |
up vote
1
down vote
If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.
Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$
$sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.
$(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$
What is the length of this interval?
answered Nov 29 at 4:17
Shubham Johri
3,084413
add a comment |
up vote
1
down vote
If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.
Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$
$sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.
$(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$
What is the length of this interval?
answered Nov 29 at 4:17
Shubham Johri
3,084413
add a comment |
up vote
1
down vote
up vote
1
down vote
If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.
Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$
$sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.
$(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$
What is the length of this interval?
answered Nov 29 at 4:17
Shubham Johri
3,084413
If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.
Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$
$sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.
$(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$
What is the length of this interval?
answered Nov 29 at 4:17
Shubham Johri
3,084413
answered Nov 29 at 4:17
Shubham Johri
3,084413
answered Nov 29 at 4:17
Shubham Johri
3,084413
answered Nov 29 at 4:17
Shubham Johri
3,084413
3,084413
add a comment |
add a comment |
up vote
0
down vote
Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
add a comment |
up vote
0
down vote
Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
answered Nov 29 at 8:16
Mostafa Ayaz
13.6k3836
13.6k3836
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%2f3018138%2fdetermine-whether-the-sequence-sin-fracn-pi2-converges-and-prove-your-ans%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
Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 at 3:44
1
The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 at 3:47