Fake proof, and a claim!

Question

It is well-known that $$\sum_{n=1}^{\infty}(-1)^n$$ diverges, because $\lim_{n \to \infty}a_n\neq0$ But look at this as a fake proof. Someone claimed that he can show $$s=-1+1-1+1-1+1+\cdots \sim \log 2$$ I Know that is $\color{red} {\bf \text{fake}}$, but I am looking for the idea behind his claim.

I tried to $s=1,s=-1 ,s=\frac12$ but I can't find some more idea. Again I say, I know that is $\color{red} {\bf \text{fake}}$. but may be an fantastic idea behind his claim.

for example my trial was like below $$s=(-1+1)+(-1+1)+(-1+1)+...=0\\$$simplify the parentheses $$s=-1+(1-1)+(1-1)+(1-1)+(1-1)+...=-1\\$$ rearrange and simplify parentheses or pick up one $+1$ and simplify parentheses $$s=+1+(1-1)+(1-1)+(1-1)+(1-1)+...=1$$ or $$s=(-1+1)+(-1+1)+(-1+1)+...=\\-1-\underbrace{(-1+1)-(-1+1)-(-1+1)...}_{-s}\\\to \\s=-1+s \\s=\frac12$$ Now :is there any idea to show $s\to $ any other number ?
for example can you show $s\to \frac{3}{$}$ ?

I think It is necessary to see some $\color{red} {\bf \text{fake}}$ proof like this , because some student or others may ask about them .

Thanks for any hint in advance.

Answer 1

This is all fake because you are manipulating a divergent series as if it were convergent. If we are allowed this, we can show that $\sum_{n=1}^\infty (-1)^n$ is equal to any real number.

Indeed, suppose $a$ is a real number. Then $$ \begin{align*} \sum_{n=1}^\infty (-1)^n &= -1 + 1 -1+1-1+1-+\cdots \\ &= (-1+1)+(-1+1)+\cdots \\ &= 0 \\ &= 0a \\ &= ((-1+1)+(-1+1)+\cdots)a \\ &= ((1-1)+(1-1)+\cdots)a \\ &=(1+((-1+1)+(-1+1)+\cdots))a \\ &=(1+0)a \\ &=a \\ \end{align*} $$

Answer 2

In general, there is no way to relate a divergent series to just any value you want. Rather, you need to define what a divergent series is.

For a convergent infinite series, we generally agree that

$$\sum_{n=1}^\infty a_n\equiv\lim_{N\to\infty}\sum_{n=1}^N a_n$$

This has two effects: it defines what an infinite series is, and it does so uniquely. That is, from here, you can go on to prove various mathematical properties of our infinite series.

Interestingly though, if this series does not converge absolutely, there exists a rearrangement $b_n$ such that $b_n$ goes through all the natural $k\in[1,\infty)$, though not in order, and

$$\lim_{N\to\infty}\sum_{n=1}^N a_n\ne\lim_{N\to\infty}\sum_{n=1}^N a_{b_n}$$

This is known as the Riemann series theorem, which helps me stress the point that

you need to define what your gibberish means!

Or else you'll expect something to work, but it doesn't.

To emphasize, if I haven't done so enough already, just because you moved some terms around does not mean you have the original series. In order to move terms around rigorously, you should, for a convergent series, check if your rearrangements are doable within the partial sums! If you can't fit such rearrangements in the partial sums, your rearrangement is likely invalid.

In the same manner, the various definitions of what a divergent series means only assign one value to each series. For example, we could use the Cesàro summation to define

$$\sum_{n=1}^\infty a_n\equiv\lim_{N\to\infty}\frac1n\sum_{k=1}^N\sum_{n=1}^ka_n$$

Or we could use the stronger Abel summation to define

$$\sum_{n=1}^\infty a_n\equiv\lim_{x\to1^-}\lim_{N\to\infty}\sum_{n=1}^Na_nx^n$$

Or we could use a Ramanujan summation or a Borel summation or what have you, but each of these definitions only gives one value per divergent series!

That said, what does a divergent sum mean to you? If it is any of the above, you will find that

$$\sum_{n=1}^\infty(-1)^n=-\frac12$$

Answer 3

I gather that you're asking if it's possible, for any $x\in\mathbb R$, to construct an incorrect proof that the series has value $x$. The trivial answer is yes: Incorrect proofs can prove anything! But here's a more concrete idea...

First of all, the following is a true theorem:

• For any $x\in\mathbb R$, there is a series of $\pm 1$ terms whose Cesaro sum is $x$.

Now, the Cesaro sum of the original series $\sum_{n=1}^{\infty}(-1)^n$ is $-\frac12$. That much is unambiguous. But if you tacitly imply the following two falsehoods to someone:

• Rearranging a Cesaro summable series does not change the Cesaro sum. (Wrong: It often does.)
• The Cesaro sum of a series is the same as the ordinary sum. (Wrong: It's different.)

Then you can convince them that the sum is $x$.