What does 1 - 1 + 1 - 1 + 1... equal?
This is part of a series on common misconceptions.
Is this true or false?
\[1-1+1-1+1-1+ \dots = 0\]
Why some people say it's true: If the terms are grouped like this \((1-1)\) , then the sum obviously becomes \(0\) .
Why some people say it's false: If the terms are grouped like this \(1+(-1+1) + (-1+1) \dots\) , then the sum obviously becomes \(1\) as the next terms cancel out each other.
The statement is \( \color{red}{\textbf{false}}\).
Proof:
The sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. The sequence of partial sums of above series i.e Grandi's series is 1, 0, 1, 0, ..., which clearly does not approach any number (although it does have two accumulation points at 0 and 1). Therefore, Grandi's series is divergent and its sum cannot have a definite answer.
Rebuttal: Since the sum is of a series of integers , the answer must be an integer.Reply: There is no finite answer to the sum. By divergent test , the series fails to converge and hence there is no definite answer to this sum of series. Strange things can happen at infinity. For example, we can have the sum of infinitely many rational numbers as irrational:
\[ \pi = 3 + 0.1 + 0.04 + 0.005 + 0.0009 + \ldots \]
Rebuttal: We observe that \(1-1+1-1 \dots\) is an infinite geometric progression with first term \(a=1\) and common ratio \(r=-1\) and hence the sum is given by \(S=\dfrac{a}{1-r} = \dfrac{1}{1-(-1)} = \dfrac{1}{1+1} = \dfrac{1}{2}\)
Reply: The sum of geometric progression only converges if \(-1<r<1\) in this case, \(r = -1\) , which is not in the required interval. Other than this , there is a way to extend the idea of this summation, either through cesaro summation or an analytic method. However, if this was the case, then it should be specificed in the problem.
See Also