Showing that a piece-wise defined sequence converges to $0$ via $epsilon$-definition?
I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.
Does it suffice to consider the the odd/even terms of the sequence separately? e.g.
If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$
And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.
real-analysis sequences-and-series
asked Mar 5 '18 at 17:41
Ryan
399211
|
show 3 more comments
I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.
Does it suffice to consider the the odd/even terms of the sequence separately? e.g.
If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$
And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.
real-analysis sequences-and-series
asked Mar 5 '18 at 17:41
Ryan
399211
1
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56
|
show 3 more comments
I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.
Does it suffice to consider the the odd/even terms of the sequence separately? e.g.
If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$
And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.
real-analysis sequences-and-series
asked Mar 5 '18 at 17:41
Ryan
399211
I wish to show if $S_n = begin{cases}
2e^{-n}, & mbox{if n is even } \
-frac{3}{n}, & mbox{if n is odd }
end{cases}$ then we have $S_n to 0$ as $n to infty$.
Does it suffice to consider the the odd/even terms of the sequence separately? e.g.
If $n$ is even, take $ N > log(2/epsilon)$. Then $n > N implies n > log(2/3) implies |2e^{-n}|< epsilon$
And if $n$ is odd, then take $N > -frac{3}{epsilon}$. Then $n> N implies n > -frac{3}{epsilon} implies |-frac{3}{n}| < epsilon$.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Mar 5 '18 at 17:41
Ryan
399211
asked Mar 5 '18 at 17:41
Ryan
399211
edited Mar 5 '18 at 17:54
asked Mar 5 '18 at 17:41
Ryan
399211
asked Mar 5 '18 at 17:41
Ryan
399211
asked Mar 5 '18 at 17:41
Ryan
399211
399211
1
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56
|
show 3 more comments
1
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56
1
1
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56
|
show 3 more comments
1 Answer
1
active
oldest
votes
That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
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%2f2678046%2fshowing-that-a-piece-wise-defined-sequence-converges-to-0-via-epsilon-defin%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
That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
add a comment |
That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
add a comment |
That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
That's right. Also notice that for $n$ being odd, we must have $$N>{3over epsilon}$$not $N>-{3over epsilon}$. The general $N$ for both even and odd terms then would become$$N>maxleft{{3over epsilon} , log{2over epsilon} right}$$
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
answered Dec 1 '18 at 14:10
Mostafa Ayaz
13.6k3836
answered Dec 1 '18 at 14:10
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%2f2678046%2fshowing-that-a-piece-wise-defined-sequence-converges-to-0-via-epsilon-defin%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
1
You need to be more careful. $N$ is supposed to be an (positive) integer and neither of your choices are integers and one of them is even negative!
– LoveTooNap29
Mar 5 '18 at 17:51
Your first $N$ is not a function of $epsilon$ so something is wrong.
– Jair Taylor
Mar 5 '18 at 17:53
In general, the trick in this kind of situation is to find $N_1$ for the first situation and $N_2$ for the second situation and then set $N = max(N_1,N_2)$. Then if $n > N$ then $n > N_1$ and $n > N_2$ so everything works.
– Jair Taylor
Mar 5 '18 at 17:54
But there has to be just one N for each given $epsilon$.
– Jair Taylor
Mar 5 '18 at 17:55
Thanks for the reply @LoveTooNap29 Instead of saying "take $N= ldots $" would it be correct to say "take $N > ldots $"?
– Ryan
Mar 5 '18 at 17:56