What we get if we add 1/2 infinite times

Question

What we get if we add 1/2 infinite times

346 Views Asked by Bumbble Comm At 03 Apr 2026 - 7:46

I want to know if this is correct

We have this sums:

$$S1=1-1+1-1+1-1+1-1+1-1...=\frac12$$

$$S2=1-2+3-4+5-6+7-8...=\frac14$$

$$S3=1+2+3+4+5+6+7+8...=-\frac{1}{12}$$

If we take

$$S4=S1+S1+S1+S1+...=\frac12+\frac12+\frac12+\frac12...$$

we get

S4= +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 +1 -1 … +1 -1 +1 -1 … +1 -1 … + . . . +1 -1 +2 -2 +3 -3 +4 -4 +5 -5 +6 -6 +7 -7 +8 -8 …

we find that

$$S4=S3-S3=(1+2+3+4+5+6+7+8...)-(1+2+3+4+5+6+7+8...)=1-1+2-2+3-3+4-4+5-5+6-6+7-7+8-8...=\left(-\frac{1}{12}\right)-\left(-\frac{1}{12}\right)=-\frac{1}{12}+\frac{1}{12}=0$$

therefore

$$S4=S1+S1+S1+S1⋯=\frac12+\frac12+\frac12+\frac12...=0$$

And also, if we take:

$$S5=S2+S2+S2+S2...=\frac14+\frac14+\frac14+\frac14+...$$

we get

S5= +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 +15 -16 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 +15 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 -14 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 +13 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 -12 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 +11 … +1 -2 +3 -4 +5 -6 +7 -8 +9 -10 … +1 -2 +3 -4 +5 -6 +7 -8 +9 … +1 -2 +3 -4 +5 -6 +7 -8 … +1 -2 +3 -4 +5 -6 +7 … +1 -2 +3 -4 +5 -6 … +1 -2 +3 -4 +5 … +1 -2 +3 -4 … +1 -2 +3 … +1 -2 … +1 … + . . . = +1 -1 +2 -2 +3 -3 +4 -4 +5 -5 +6 -6 +7 -7 +8 -8 …

we find that

$$S5=S3-S3=(1+2+3+4+5+6+7+8...)-(1+2+3+4+5+6+7+8...)=1-1+2-2+3-3+4-4+5-5+6-6+7-7+8-8...=\left(-\frac{1}{12}\right)-\left(-\frac{1}{12}\right)=-\frac{1}{12}+\frac{1}{12}=0$$

therefore

$$S5=S2+S2+S2+S2...=\frac14+\frac14+\frac14+\frac14...=0$$

also

$$S5=S4$$

Original Q&A

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**Bumbble Comm** · Answer 1 · 2015-06-28 15:47:24

Bumbble Comm On 28 Jun 2015 - 3:47

You start with wrong conditions.

For example, the infinite sum $\sum_n^{\infty} (-1)^n $ is not equal to $1/2$.

**Bumbble Comm** · Answer 2 · 2015-06-28 16:03:53

Every claim you started with is wrong. Your $S_{1}, S_{2}, S_{3}$ are all divergent series.

It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value.

**Bumbble Comm** · Answer 3 · 2015-06-28 16:17:22

As the other answers have said, your reasoning is very off: these are divergent sequences, and don't behave the way you think they do.

On the other hand, a reasonable question to ask is: "Can we make sense of these divergent series?" The answer is yes, although it is not easy: there are several methods for summing (some) classically divergent series. See https://en.wikipedia.org/wiki/Cesàro_summation (according to which which $1-1+1-1 . . .={1\over 2}$, https://en.wikipedia.org/wiki/Borel_summation, or https://en.wikipedia.org/wiki/Divergent_series#Abel_summation.

As for the rest of your reasoning, you rely on properties of series which do not hold for divergent series, even when interpreted according to Cesaro or other - e.g., being able to freely rearrange and combine terms. So to be frank, your reasoning is far off. But there are very interesting things you can do with divergent series!

(An example of the bizarre properties of divergent series: according to Cesaro summation, $-1+1-1+1. . .=-{1\over 2}$, even though it's "clearly" the same as $1-1+1-1.. .$.)

**Bumbble Comm** · Answer 4 · 2015-06-28 16:18:27

There is no official way to add these divergent alternating series, but $\sum (-1)^n$ is known as Grandi's series:

$$ \sum_{n=0}^\infty (-1)^n = \begin{cases} 1 & \text{ if }n\text{ is odd} \\ 0 & \text{ if }n\text{ is even} \\ \end{cases} \approx \frac{1}{2}$$

And for your second series $(1 - 1 + 1 - 1 + \dots)^2 = 1 - 2 + 3 - 4 + \dots = \frac{1}{4}$ Wikipedia gives this nice figure

Doesn't it seem obvious that $S5 = \frac{1}{2} S4$ ? However wikipedia gives the value $\frac{1}{2}$.

**Bumbble Comm** · Answer 5 · 2015-06-28 16:20:59

I think it is praiseworthy that you are interested in these infinite sums and are thinking of ways to combine them in order derive new results. However there is a problem: all these sums are not strictly convergent, but exist only if one extends a correct result to a case where they are not really properly defined. Unfortunately this means that the operations that you carry out are also dubious.

I suggest that you start with strictly converging sums. For example your $S4$ is best written as:

$$S4(t) = 1 - t + t^2 - t^3 + t^4 - t^5 + t^6-t^7+t^8-t^9+t^{10}........$$

where t is a positive real number smaller than 1. You can now consider what happens if you calculate $S4(t) + t^2S4(t) + t^4S4(t) + ....$ as in your example.

What we get if we add 1/2 infinite times

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