what is 1+1-1+1-1+1...?

👨🏻‍🏫 學術台數學

[匿名]

2008-07-16 10:00 am

更新1:

the question is about infinite series. please give a quick proof or reason why.

更新2:

the nth term is (-1)^(n), except the first term, so the next number in the series is -1.

回答 (21)

[匿名]

2008-07-16 10:33 am

✔ 最佳答案

The series is not absolutely convergent since the modulus of each term is 1 and the sum to infinity of 1 is divergent.

This shows that the order in which you add terms does matter, for example:

(1+1)-(1+1)-(1+1)-... = 2-2-2-... -> - infinity

1+(1-1)+(1-1)+(1-1)+(1-1)... = 1

(1+1) -1 + 1 - 1 ... = 2 -1+1-1+1...=2

Also using sum to infinity of geometric series formula (which is not necessarily valid since the ratio is not strictly between -1 and 1);

sum to infinity = a/(1-r) = 1/(1-(-1)) = 1/2

So the answer is, there is no real answer unless an order is specified.

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[匿名]

2008-07-16 5:48 pm

1+1-1+1-1+1… = 1 + Î£[n=0 to n= infinity] (-1)^n
S(m) = Î£[n=0 to n=m] (-1)^n is a partial sum of series
S(0) = 1
S(1) = 0
S(2) = 1
S(3) = 0 ….

The series of partial sums 1, 0, 1, 0, … don’t have a finite limit.
=> 1 + Î£[n=0 to n= infinity] (-1)^n is a divergent series.

參考: http://en.wikipedia.org/wiki/Alternating_series

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[匿名]

2008-07-16 5:19 pm

/
2

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bskelkar

2008-07-16 5:14 pm

1+1-1+1-1+1 or 1+1-1+1-1+1+....or1+1-1+1-1+1-....?
You see, unless you give in formula form, the n th term cannot be guessed from first few terms.
You must give it in the form 1 +(-1)^(n+1) or like.

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[匿名]

2008-07-16 5:10 pm

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An ESL Learner

2008-07-16 5:08 pm

1 + 1 - 1 + 1 - 1 + 1...
= 2 - 1 + 1 - 1 + 1...
= 1 + 1 - 1 + 1...
= 2 - 1 + 1...
= 1 + 1...
= 2...

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[匿名]

2008-07-16 5:07 pm

2 ?

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[匿名]

2008-07-16 5:04 pm

2 or 1

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[匿名]

2008-07-16 5:03 pm

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[匿名]

2008-07-16 5:03 pm

it's 2

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