<s>[INST] $2 - 4 + 6 - 8 + \cdots = 1 - 1 + 1 -1 + \cdots$? I want to talk about the weirdness of $2\sum\limits_{n=1}^\infty n(-1)^{n-1}$ ,
$$\sum\limits_{n=1}^\infty n(-1)^{n-1} = 1 - 2 + 3 - 4 + 5 - 6 + \dots$$
$$\times 2 \implies 2\sum\limits_{n=1}^\infty n(-1)^{n-1} = 2 - 4 + 6 - 8 + 10 - 12 + \dots\quad\longleftarrow (1)$$

But

$2\sum\limits_{n=1}^\infty n(-1)^{n-1}$
$=\sum\limits_{n=1}^\infty n(-1)^{n-1} + \sum\limits_{n=1}^\infty n(-1)^{n-1}$
$= 1 - 2 + 3 - 4 + 5 - 6 + \dots$
$\quad\quad+ 1 - 2 + 3 - 4 + 5 - 6 +\dots$
$= 1 - 1 + 1 - 1 + 1 - 1 + 1 - \dots \quad\longleftarrow (2)$

Is it true that these series are equal? The manipulation that I've made to get $(2)$ seems very unnatural but in my defense, the way I add them shouldn't matter because the result should be the same. For example, using the same manipulation on a finite series, $2( 1 + 2 + 3) = 1 + 3 + 5 +3$. It yields the same sum but I'm not sure if this is applicable to infinite series a well.

Using $S_\infty = \large \frac{a}{1-r}$ on $(2)$, it can be seen that it's sum is $\large \frac{1}{2}$. Can I say that the sum of $(1)$ is the same?
I'm a high school student and I have no real background in sequences, so a simple and understandable answer is greatly appreciated. I may not understand many complicated terms but I am willing to learn.
 [/INST] Sangchul Lee commented above that the manipulations are correct for Abel sums.
In fact, the series $1-2+3-4+\cdots$ corresponds to the function
$$
f_1(x)=\frac{x}{(1+x)^2}=x-2x^2+3x^3-4x^4+\cdots~,
$$
the series $0+1-2+3-4+\cdots$ corresponds to the function
$$
f_2(x)=\frac{x^2}{(1+x)^2}=x^2-2x^3+3x^4-4x^5+\cdots
$$
and the series $1-1+1-1+\cdots$ corresponds to the function
$$
f_3(x)=\frac{x}{1+x}=x-x^2+x^3-x^4+\cdots~,
$$
which are convergent for $|x|<1$.
The manipulations correspond to the identity
$$
f_1(x)+f_2(x)=f_3(x)~,
$$
that is,
$$
\frac{x}{(1+x)^2}+\frac{x^2}{(1+x)^2}=\frac{x}{1+x}~,
$$
which is also valid in the limit as $x\to 1^-$.
The Abel sum of $1-2+3-4+\cdots$ is defined by
$\lim_{x\to 1^-} f_1(x)=1/4$,
the Abel sum of $0+1-2+3-4+\cdots$ is defined by
$\lim_{x\to 1^-} f_2(x)=1/4$,
and the Abel sum of $1-1+1-1+\cdots$ is defined by
$\lim_{x\to 1^-} f_3(x)=1/2$.
</s>