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I was looking into a Numberphile video here. The guy says he was unable to find an intuition. Does there exist one? Is the premise, $1-1+1-1+...=\frac{1}{2}$, reasonable mathematically?

math
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3 Answers3

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The intuition is that $\zeta(s) \equiv \sum_{n=1}^\infty n^{-s}$ evaluates to $1+2+3+\cdots$ when $s=-1$.

However, the fact of the matter is that this definition of the zeta function holds for complex $s$ such that $\textrm{Re}(s) > 1$, so it is not generally true for $s=-1$.

Now, it is possible through a process known as analytic continuation to "extend" the zeta function to $s=-1$, however when doing so, the equivalence between $\zeta(-1)$ and $1+2+3+\cdots$ breaks down.

This is because infinite series are defined as the limits of their partial sums. So $\sum_{k=1}^\infty k$ diverges, and while you can interpret it some other ways, if you want to speak the language of infinite sums, you have to accept the partial sum limit definition.

Emily
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No it is not. the sum $1 - 1 + 1 - ...$ does not converge, it has no meaning to say that since you can stop at zero or at $1$, then it equals $\frac{1}{2}$

Sometimes this stuff can have meaning but you have to define precisely what you mean by "sum", or "equal", because with stuff that goes to infinity there are a lot of problems :-D

An example I can think of is $$1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$

This series converges (so that "equal" has meaning) only if $|x| < 1$. In some areas of math people extended the above definition so that it make sense for every x, also, say, $x=3$. But it's a definition and you have to be really cautios with that.

But of course people will start telling that it is "proven" that $$1 + 3 + 9 + 27 + ... = -\frac{1}{2}$$

Ant
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I think this should have been included for the premise in the video:
Let $x:=1-1+1-1+1-\dots$, then we have $$x=1-(1-1+1-1+\dots)=1-x$$ so $2x=1,\ \ x=1/2$.

Berci
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