Mixing
Nature of the series $sum 1+(-1)^{n+1} (2n+1)$ .
$begingroup$
The series $sum 1+(-1)^{n+1} (2n+1)$ .is
1. Convergent
2. Oscillates finitely
3. Divergent
4. Oscillates infinitely
I found first few terms of this series, which are 4-4+8-8+...
So it seems like I will get such pairs if I expand the series more. But what can we conclude about the nature of the series at infinity? The series is oscillating infinitely. So can I say it is divergent?
sequences-and-series convergence
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
$endgroup$
|
show 1 more comment
$begingroup$
The series $sum 1+(-1)^{n+1} (2n+1)$ .is
1. Convergent
2. Oscillates finitely
3. Divergent
4. Oscillates infinitely
I found first few terms of this series, which are 4-4+8-8+...
So it seems like I will get such pairs if I expand the series more. But what can we conclude about the nature of the series at infinity? The series is oscillating infinitely. So can I say it is divergent?
sequences-and-series convergence
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
$endgroup$
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
1
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25
|
show 1 more comment
$begingroup$
The series $sum 1+(-1)^{n+1} (2n+1)$ .is
1. Convergent
2. Oscillates finitely
3. Divergent
4. Oscillates infinitely
I found first few terms of this series, which are 4-4+8-8+...
So it seems like I will get such pairs if I expand the series more. But what can we conclude about the nature of the series at infinity? The series is oscillating infinitely. So can I say it is divergent?
sequences-and-series convergence
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
$endgroup$
The series $sum 1+(-1)^{n+1} (2n+1)$ .is
1. Convergent
2. Oscillates finitely
3. Divergent
4. Oscillates infinitely
I found first few terms of this series, which are 4-4+8-8+...
So it seems like I will get such pairs if I expand the series more. But what can we conclude about the nature of the series at infinity? The series is oscillating infinitely. So can I say it is divergent?
sequences-and-series convergence
sequences-and-series convergence
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
edited Jan 9 at 15:24
Martin Sleziak
44.8k9118272
44.8k9118272
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
asked Jan 9 at 12:24
MathsaddictMathsaddict
3459
3459
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
1
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25
|
show 1 more comment
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
1
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
1
1
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $sum a_n $ is divergent.
answered Jan 9 at 12:28
FredFred
45.5k1848
$endgroup$
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
add a comment |
$begingroup$
Surely oscillates infinitely since $$a_{n}=2n+2quad,quad ntext{ is odd}\a_{n}=-2nquad,quad ntext{ is even}$$
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
Suppose for contradiction that the series $sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)to 0,;;text{ as }ntoinfty.$$
However, begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=begin{cases}2n+2,&text{if};n;text{is even,}\-2n,&text{if};n;text{is odd.}end{cases} end{align}
begin{align}text{Thus, for odd} ; ninBbb{N},; c_n= -2nto -infty,;text{while}; c_n= 2n+2to infty;text{for even} ;n,text{contradiction.}end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
begin{align}infty= sum^{infty}_{n=1} nleqsum^{infty}_{n=1} (2n+2)leq infty;text{and} ;-inftyleqsum^{infty}_{n=1} (-2n)leqsum^{infty}_{n=1} -n= -inftyend{align}
Option $(3)$ is eliminated because
begin{align}sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=infty; text{if};n;text{is even}end{align}
begin{align}text{and};sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-infty; text{if};n;text{is odd.}end{align}
Hence, it oscillates infinitely.
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
$endgroup$
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%2f3067393%2fnature-of-the-series-sum-1-1n1-2n1%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
$begingroup$
If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $sum a_n $ is divergent.
answered Jan 9 at 12:28
FredFred
45.5k1848
$endgroup$
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
add a comment |
$begingroup$
If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $sum a_n $ is divergent.
answered Jan 9 at 12:28
FredFred
45.5k1848
$endgroup$
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
add a comment |
$begingroup$
If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $sum a_n $ is divergent.
answered Jan 9 at 12:28
FredFred
45.5k1848
$endgroup$
If $a_n=1+(-1)^{n+1}(2n+1)$, the $(a_n)$ is unbounded, hence $(a_n)$ does not converge to $0$. Thus $sum a_n $ is divergent.
answered Jan 9 at 12:28
FredFred
45.5k1848
answered Jan 9 at 12:28
FredFred
45.5k1848
answered Jan 9 at 12:28
FredFred
45.5k1848
answered Jan 9 at 12:28
FredFred
45.5k1848
45.5k1848
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
add a comment |
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
$begingroup$
I am a little confused! Definition of a divergent series says that a series is divergent if it is not convergent. I read somewhere, bieng divergent and bieng infitely oscillatory are different. A series is said to be divergent if it diverges to infinite or -infinite.
$endgroup$
– Mathsaddict
Jan 9 at 12:57
add a comment |
$begingroup$
Surely oscillates infinitely since $$a_{n}=2n+2quad,quad ntext{ is odd}\a_{n}=-2nquad,quad ntext{ is even}$$
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
Surely oscillates infinitely since $$a_{n}=2n+2quad,quad ntext{ is odd}\a_{n}=-2nquad,quad ntext{ is even}$$
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
Surely oscillates infinitely since $$a_{n}=2n+2quad,quad ntext{ is odd}\a_{n}=-2nquad,quad ntext{ is even}$$
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
Surely oscillates infinitely since $$a_{n}=2n+2quad,quad ntext{ is odd}\a_{n}=-2nquad,quad ntext{ is even}$$
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 12:45
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
add a comment |
add a comment |
$begingroup$
For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
For a series to converge the general term must tend to $0$. In this case it doesn't so the series is divergent.
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 12:29
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
57.5k42160
add a comment |
add a comment |
$begingroup$
Suppose for contradiction that the series $sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)to 0,;;text{ as }ntoinfty.$$
However, begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=begin{cases}2n+2,&text{if};n;text{is even,}\-2n,&text{if};n;text{is odd.}end{cases} end{align}
begin{align}text{Thus, for odd} ; ninBbb{N},; c_n= -2nto -infty,;text{while}; c_n= 2n+2to infty;text{for even} ;n,text{contradiction.}end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
begin{align}infty= sum^{infty}_{n=1} nleqsum^{infty}_{n=1} (2n+2)leq infty;text{and} ;-inftyleqsum^{infty}_{n=1} (-2n)leqsum^{infty}_{n=1} -n= -inftyend{align}
Option $(3)$ is eliminated because
begin{align}sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=infty; text{if};n;text{is even}end{align}
begin{align}text{and};sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-infty; text{if};n;text{is odd.}end{align}
Hence, it oscillates infinitely.
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
$endgroup$
add a comment |
$begingroup$
Suppose for contradiction that the series $sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)to 0,;;text{ as }ntoinfty.$$
However, begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=begin{cases}2n+2,&text{if};n;text{is even,}\-2n,&text{if};n;text{is odd.}end{cases} end{align}
begin{align}text{Thus, for odd} ; ninBbb{N},; c_n= -2nto -infty,;text{while}; c_n= 2n+2to infty;text{for even} ;n,text{contradiction.}end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
begin{align}infty= sum^{infty}_{n=1} nleqsum^{infty}_{n=1} (2n+2)leq infty;text{and} ;-inftyleqsum^{infty}_{n=1} (-2n)leqsum^{infty}_{n=1} -n= -inftyend{align}
Option $(3)$ is eliminated because
begin{align}sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=infty; text{if};n;text{is even}end{align}
begin{align}text{and};sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-infty; text{if};n;text{is odd.}end{align}
Hence, it oscillates infinitely.
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
$endgroup$
add a comment |
$begingroup$
Suppose for contradiction that the series $sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)to 0,;;text{ as }ntoinfty.$$
However, begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=begin{cases}2n+2,&text{if};n;text{is even,}\-2n,&text{if};n;text{is odd.}end{cases} end{align}
begin{align}text{Thus, for odd} ; ninBbb{N},; c_n= -2nto -infty,;text{while}; c_n= 2n+2to infty;text{for even} ;n,text{contradiction.}end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
begin{align}infty= sum^{infty}_{n=1} nleqsum^{infty}_{n=1} (2n+2)leq infty;text{and} ;-inftyleqsum^{infty}_{n=1} (-2n)leqsum^{infty}_{n=1} -n= -inftyend{align}
Option $(3)$ is eliminated because
begin{align}sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=infty; text{if};n;text{is even}end{align}
begin{align}text{and};sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-infty; text{if};n;text{is odd.}end{align}
Hence, it oscillates infinitely.
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
$endgroup$
Suppose for contradiction that the series $sum 1+(-1)^{n+1} (2n+1)$ converges, then the sequence $$1+(-1)^{n+1} (2n+1)to 0,;;text{ as }ntoinfty.$$
However, begin{align} c_n:= 1+(-1)^{n+1} (2n+1)=begin{cases}2n+2,&text{if};n;text{is even,}\-2n,&text{if};n;text{is odd.}end{cases} end{align}
begin{align}text{Thus, for odd} ; ninBbb{N},; c_n= -2nto -infty,;text{while}; c_n= 2n+2to infty;text{for even} ;n,text{contradiction.}end{align}
Therefore, option $(1)$ is eliminated. Now, by Direct Comparison
begin{align}infty= sum^{infty}_{n=1} nleqsum^{infty}_{n=1} (2n+2)leq infty;text{and} ;-inftyleqsum^{infty}_{n=1} (-2n)leqsum^{infty}_{n=1} -n= -inftyend{align}
Option $(3)$ is eliminated because
begin{align}sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=infty; text{if};n;text{is even}end{align}
begin{align}text{and};sum^{infty}_{n=1} 1+(-1)^{n+1} (2n+1)=-infty; text{if};n;text{is odd.}end{align}
Hence, it oscillates infinitely.
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
edited Jan 9 at 14:37
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
answered Jan 9 at 13:44
Omojola MichealOmojola Micheal
1,851324
1,851324
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%2f3067393%2fnature-of-the-series-sum-1-1n1-2n1%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
$begingroup$
You can easily rule out $(1)$ and $(3)$. Now look at the positive terms to find the final answer.
$endgroup$
– Peter
Jan 9 at 12:29
$begingroup$
@Peter, do you mean (1) and (2) rather than (1) and (3)?
$endgroup$
– Barry Cipra
Jan 9 at 12:57
$begingroup$
@BarryCipra I mean $(3)$ Divergent means that the partial sums exceed every positive (or in absolute terms every negative) value from some point on. Here we have the partial sums $4,0,8,0,12,0,16,0,cdots$, so the series does not diverge.
$endgroup$
– Peter
Jan 9 at 12:58
$begingroup$
@Peter, I grew up with the definition that divergent simply means not convergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:02
1
$begingroup$
@Peter, it sounds like the modifier bestimmt (which translates as "certainly") is important here. In English, we would say something like "diverges to infinity" (or even "converges to infinity," if you get fancy by compactifying the reals with an extra point or two). Maybe the key thing here is for the OP to stipulate their understanding of what it means to be divergent.
$endgroup$
– Barry Cipra
Jan 9 at 13:25