@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.