I've been wondering what is the solution of:
∞
∑ 1/(-1^n)
n=1
∞
∑ 1/(-1^n)
n=1
@IsaiahCui said in #1:
> I've been wondering what is the solution of:
>
> ∞
> ∑ 1/(-1^n)
> n=1
Do you consider infinty odd or even?
> I've been wondering what is the solution of:
>
> ∞
> ∑ 1/(-1^n)
> n=1
Do you consider infinty odd or even?
The series does not converge, it has no limit in the real numbers. You're welcome.
Best,
Lichess Mathematical Service.
Best,
Lichess Mathematical Service.
@IsaiahCui said in #1:
> I've been wondering what is the solution of:
>
> ∞
> ∑ 1/(-1^n)
> n=1
I would like to add a few details to the answer given by @m011235.
I suppose that you ask about the sum of the series sum_{n=1}^{infinity}1/(-1)^{n}.
Since 1/(-1)^{n}=(-1)^{n} for every natural n, we have
sum_{n=1}^{infinity}1/(-1)^{n}=sum_{n=1}^{infinity}(-1)^{n}.
The series sum_{n=1}^{infinity}(-1)^{n} does not have the sum. Consequently
it is not convergent.
Formally the term 1/(-1^n) should be treated as 1/(-(1^{n})), that is 1/(-1)
which is equal to -1. But then this term could be simplified to the term which
does not depend on n, so I suppose that your intention was different.
However, formally, the sum of the series sum_{n=1}^{infinity}1/(-1^n)
is equal to the sum of the series sum_{n=1}^{infinity}(-1) which is equal to
-infinity.
> I've been wondering what is the solution of:
>
> ∞
> ∑ 1/(-1^n)
> n=1
I would like to add a few details to the answer given by @m011235.
I suppose that you ask about the sum of the series sum_{n=1}^{infinity}1/(-1)^{n}.
Since 1/(-1)^{n}=(-1)^{n} for every natural n, we have
sum_{n=1}^{infinity}1/(-1)^{n}=sum_{n=1}^{infinity}(-1)^{n}.
The series sum_{n=1}^{infinity}(-1)^{n} does not have the sum. Consequently
it is not convergent.
Formally the term 1/(-1^n) should be treated as 1/(-(1^{n})), that is 1/(-1)
which is equal to -1. But then this term could be simplified to the term which
does not depend on n, so I suppose that your intention was different.
However, formally, the sum of the series sum_{n=1}^{infinity}1/(-1^n)
is equal to the sum of the series sum_{n=1}^{infinity}(-1) which is equal to
-infinity.
Lichess create another topic called as Off Mathematics Discussions
@Marcin2 said in #5:
> I would like to add a few details to the answer given by @m011235.
> I suppose that you ask about the sum of the series sum_{n=1}^{infinity}1/(-1)^{n}.
> Since 1/(-1)^{n}=(-1)^{n} for every natural n, we have
> sum_{n=1}^{infinity}1/(-1)^{n}=sum_{n=1}^{infinity}(-1)^{n}.
> The series sum_{n=1}^{infinity}(-1)^{n} does not have the sum. Consequently
> it is not convergent.
> Formally the term 1/(-1^n) should be treated as 1/(-(1^{n})), that is 1/(-1)
> which is equal to -1. But then this term could be simplified to the term which
> does not depend on n, so I suppose that your intention was different.
How’d u calculate that?
> However, formally, the sum of the series sum_{n=1}^{infinity}1/(-1^n)
> is equal to the sum of the series sum_{n=1}^{infinity}(-1) which is equal to
> -infinity.
> I would like to add a few details to the answer given by @m011235.
> I suppose that you ask about the sum of the series sum_{n=1}^{infinity}1/(-1)^{n}.
> Since 1/(-1)^{n}=(-1)^{n} for every natural n, we have
> sum_{n=1}^{infinity}1/(-1)^{n}=sum_{n=1}^{infinity}(-1)^{n}.
> The series sum_{n=1}^{infinity}(-1)^{n} does not have the sum. Consequently
> it is not convergent.
> Formally the term 1/(-1^n) should be treated as 1/(-(1^{n})), that is 1/(-1)
> which is equal to -1. But then this term could be simplified to the term which
> does not depend on n, so I suppose that your intention was different.
How’d u calculate that?
> However, formally, the sum of the series sum_{n=1}^{infinity}1/(-1^n)
> is equal to the sum of the series sum_{n=1}^{infinity}(-1) which is equal to
> -infinity.
Google needs to create a New Mathematical Tab
There are three types of people in this world.
Those who are good at math.
And those who aren't.
Those who are good at math.
And those who aren't.
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