Why isn't $\sqrt{n}$ always 1?

Question

My friend recently came to me with this "proof":

$ a = \sum_0^\infty (-1)^n \times 1 = 1 - 1 + 1 - 1 + \dots = \frac{1}{2} $

$ \sqrt{n} = n^{\frac{1}{2}} = n^{1 - 1 + 1 - 1 + \dots} = n^{(1 - 1) + (1 - 1) + \dots} = n^{(1 - 1)} \times n^{(1 - 1)} \times \dots = n^0 \times n^0 \times \dots = 1 \times 1 \times 1 \dots = 1$

What's the problem here?

I have two ideas myself: a. that $1 - 1 + 1 - \dots$ is just an idea and not something that can be adopted to "usual" calculations (if so, how are these things called?). Or b., the error lies in finally "deciding" the previously "undecided" value of $a$ by favouring a certain bracketing-order? He could have chosen $1 + (1 - 1) + (1 - 1) + \dots = 1 + (0) + (0) + \dots = 1$ instead at that point.

Or is it something completely different? Clever and funny idea though.

Answer 1

If you could use your argument, then:

$$\frac{1}{2}=1+(-1)+1+(-1)+\cdots = (1-1)+(1-1)+\cdots =0$$

So you hardly need to go to square roots to find a contradiction.

You are dealing with a different sort of "summation" here than a standard summation. It does not allow you to group terms. In the more general sense, it is a Ramanujan summation, and we should, to be precise, write $(\mathfrak R)$ to indicate this:

$$1+(-1)+1+(-1)+\cdots = \frac{1}{2}\,(\mathfrak R)$$

(There are also some alternative summations that would return the same value, such as Cesàro summations, but Ramanujan summations is one of the most general. When Cesàro summation exists, a Ramanujan summation exists.)

Unfortunately, Ramanujan sums are deeply technical, and a lot of popular math sites and videos nowadays seems to want to shout:

"OMG, $1+2+3+\cdots =\frac{-1}{12}$!!!"

It's been a source of much frustration on this site because, while there is some real math being loosely described in the above:

1. It is an abuse of standard notation, and
2. It misleads students just learning infinite sums.

Ramanujan (and other non-standard sums) can be mysterious. For example, you can't simply remove zeros. For example:

$$1+0+(-1)+1+0+(-1)+\cdots =\frac{2}{3}\,(\mathfrak R)$$

But all we've done is add zeros between each $1,-1$ pair.

We've also seen that you can't just group terms together. (Technically, you also can't do this in standard infinite sums, either, but if a sum is already know to converge, we can group the terms like this and it still converges to the same number.)

The sorts of manipulations you can do with the terms, when computing a Ramanujan sum, are very different from (and more limited than) the sorts of manipulations you can do with standard infinite series.

Ramanujan sums can be fascinating - especially when they seem to be useful in dealing with facts from physics - but you have to be very careful with them. They are not as simple as they appear.

Answer 2

The problem is $$ u_n=\left(-1\right)^n $$ does not tend to $0$ hence the series does not exist. The value you gave it is not rigorous and has no sense at all, first what's behind $" + \ ..."$ ?

Answer 3

Right off the bat, the statement that the infinite series $1-1+1-1+...$ is equal to $1\over 2$ is nonsense. There are ways in which we can assign values to certain divergent series, including the one above, but in no sense can we actually treat the two as equal.

This is discussed in the particular case above here; more generally, look up "Cesaro summation" and "Abel summation" to get started. What these "summation" methods do is assign individual numbers to certain "formal series" (series where I'm not paying attention to whether they converge or not; really, I'm just talking about the sequences of partial sums) like $1-1+1-1+...$ above in ways which satisfy certain nice properties. Sometimes notation is horribly abused, and people write something like $$"1-1+1-1+...={1\over 2},"$$ or perhaps more internet-famously these days $$"1+2+3+4+5+...=-{1\over 12}."$$ But it must be understood that these are not actual equalities in any sense.

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I think the following example might help explain the difference between producing equalities and assigning numbers with regard to summing divergent series.

There are lots of ways to assign numbers to formal series. Some of these include:

• $\alpha$: To the formal series $s_1+s_2+s_3+...$, assign it $s_3$.

• $\beta$: To the formal series $s_1+s_2+s_3+...$, assign it its sum if it converges and otherwise assign it $17$.

And so on. Now the above are silly, and they have bad properties:

• If we replace $s_1+s_2+s_3+...$ with $0+s_1+s_2+s_3+...$, $\alpha$ changes from $s_3$ to $s_2$, even though these two formal series really "should" have equal value.

• If we consider the three formal series $$s=s_1+s_2+s_3+...,\quad t=t_1+t_2+t_3+..., \quad s+t=s_1+t_1, s_2+t_2, s_3+t_3, ...$$ $\beta$ assigns all three the value "$17$," even though the third "should" be assigned the sum of the first two.

Methods like Cesaro summation do not yield equalities; rather, they are again methods of assigning numbers to formal series. The difference between them and the pointless stuff above is that they do have various nice properties, but they are the same "type" of thing. While of mathematical value, you can't use them in the same way that you use actual sums of series.