Is the series $ 1-1+1-1+1+\cdots = -1+1-1+1-1+1-\cdots $?

Question

We have the series

$$ S = 1-1+1-1+1-1+1 \cdots $$

If we manipulate S, we get that $$ 1 - S = S $$ $$ 1 = 2S $$ $$ \frac{1}{2} = S $$

Also, if we re-order S, we get that

$$ S = (1-1)+(1-1)+(1-1)\cdots $$

$$ = -1+1-1+1-1+1-1+1\cdots $$

So $$ -S = S + 1 $$

Solving for S we get that

$$ S = -\frac{1}{2} $$

My question is, is the new S re-ordered considered the same series as S? When we can say that two series are the same in that context? Thanks.

Answer 1

First of all, this series is not convergent, so there's no such thing as sum of the series $S$.

Even if it was a convergent series, like for example $$ 1 - \frac12 + \frac13 - \frac14 + \dots $$ in general, if you rearrange the terms, it is possible to obtain a different sum. For example, if you rearrange them as follows: $$ 1 + \frac13 -\frac12 + \frac15 +\frac17-\frac14 + \frac19+\frac{1}{11}-\frac16 + \dots $$ (two positive terms then one negative, repeat) you will get a different sum.

Only if the series $\sum_{n=1}^\infty a_n$ is absolutely convergent, that is, the series $\sum_{n=1}^\infty |a_n|$ is convergent, then it is guaranteed that rearranging the terms will not change the sum. So, for example in the series $$ 1-\frac14 + \frac19-\frac{1}{16}+\frac{1}{25} + \dots$$ you can rearrange terms whatever way you want, and the sum will remain the same.

The theorem it states that is called the Riemann series theorem.

Answer 2

One of the simplest way to get this series will be with the help of Binomial Expansion of $\frac{1}{2}$ as follows

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

Expansion will be as follows

$1^{-1} + \frac{(-1)}{1!}(1)^{-2} \cdot(1)^1+\frac{(-1)\cdot(-2)}{2!}1^{-3}\cdot1^2+... = 1-1+1-1+…$

Clearly now in your question

LHS = $\frac{1}{2} $

RHS = $-1+\frac{1}{2} = \frac{-1}{2} $

In this context RHS $\neq$ LHS

Some more interesting series with Binomial Expansion will be awesome, may be sometimes divergent series also

For example

Expansion of $\frac{1}{3}$ can be done in 2 ways

Conventional and logic way of expansion will be $$\frac{1}{3} = \frac{1}{2+1 }= (2+1)^{-1}$$ which will expand into

$$ 2^{-1}+ \frac{-1}{1!}\cdot 2^{-2}+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{-3}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{-4}...$$

$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...$$

Another way to put this Binomial Expansion will be $$\frac{1}{3} = \frac{1}{1+2}= (1+2)^{-1}$$

Expansion now will give different result

$$ 1^{-1}+ \frac{-1}{1!}\cdot 1^{-2}\cdot2^1+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{2}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{3}...$$

And this will be $$1-2^1+2^2-2^3+...$$ a divergent series alternating from $-\infty$ to $+\infty$