Odd series behavior?

[bowma166]

Today, my professor said something like "The series 1 + -1 + 1 + -1 and so on is defined to be one half... but let's not go into that." and then didn't feel like explaining when people asked him why. I have no idea why that would be true...

It seems like a similar case might be

\int_{0}^{\infty}\sin x\,\textrm{d}x

but that isn't defined to be one half or zero or anything at all.

So why oh why is this true?

\sum_{n=0}^{\infty}\left(-1\right)^{n}=\frac{1}{2}

[sutupidmath]

So why oh why is this true?

\sum_{n=0}^{\infty}\left(-1\right)^{n}=\frac{1}{2}

THis is true.....because it is NOT true!

the alternating series
\sum_{n=0}^{\infty}\left(-1\right)^{n}

does not converge at all.
One will get either 1 or 0 as the final sum, depending on how you group the terms. So, by the definition of what we mean with convergence it does not converge.

BUT, i have heard of some sort of Eulers method, or Ramaujan summation, or some different kind of summation, and that might be true, but i have no knowledge whatsoever of those summations.

So, this is not true, if the summation is the common one, i don't know about the others. But people here will enlighten you, just wait until this thread catches the eyes of the right people....

[qntty]

Although the series diverges (http://en.wikipedia.org/wiki/Grandi%27s_series), there are methods of assigning a value (http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation#Examples) which yeilds 1/2

[yyat]

Another reason why one might want the sum to be 1/2 is that

\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}

for -1<x<1. If you put in x=-1, then the series on the left diverges, but on the right you get 1/2. This is basically the idea behind "Abel summation", which is a weaker form of Cesaro's method.

[Santa1]

I've usually seen Ramanujan summation being used in cases like that.

Equivalently you can write it in terms of the zeta function, by
\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = (1-2^{1-s})\zeta(s)
so that,

\sum_{n=0}^\infty (-1)^{n} = \eta(0) = -\zeta(0) = \frac{1}{2}

If you have seen the derivation of the functional equation (which is as far as I have understood is Ramanjuan summation) for the zeta function this sort of "makes sense".