TITLE: $1-2+3-4+\dots = \frac{1}{4}$
QUESTION [8 upvotes]: Recently I came across this paradoxical equation:
$$1-2+3-4+\dots = \frac{1}{4}$$
More rigorously, this infinite sum can be write as
$$\sum_{n=1}^{\infty} n(-1)^{n-1}$$
and it is well known to diverge since, for instance
$$1=1$$
$$1-2=-1$$
$$1-2+3=2$$
$$1-2+3-4=-2$$
$$1-2+3-4+5=3$$
$$1-2+3-4+5-6=-3$$
Now, the thing I cannot explain to myself is why the sum equals $\frac{1}{4}$, although it doesn't tend towards any limit. Here's because, I think, the equality above is called "paradoxical".
I (tried to) read the Wikipedia article, but it is full of material I've never encountered in my studies, full of examples and analogies that made me even more confused.
The only thing I believe to have understood concerning this equality, written for the first time by Leonhard Euler, is that the latter came to the rusult by reducing the polynomial $$1-2x+3x^2-4x^3+\dots$$ in $$\frac{1}{(1+x)^2}$$
Thus, if we take $x=1$, we obtain the equality above. But how is this reduction done?
I'm afraid if I couldn't realise the answers to this question on Wikipedia, but I just need an explanation that would make me understand, even intuitively, the paradoxical nature of the equality and its mathematical derivation by means of even one clear reasoning.

REPLY [3 votes]: The other answers are good as they more or less explain what is going on. This answer will seek to explain the $\frac{1}{4}$. Let us first look at a simpler series that exhibits similar enough behavior. That is the series, $$1-1+1-1+1-1\cdots.$$ Note that this is a divergent geometric series, but it is a a fly's hair away from being convergent. What do I mean? We can write down a power series, $$\frac{1}{1+x}=1-x+x^x-x^3+\cdots\mbox{ when } |x|<1.$$ If we blindly plug in $x=1$ (even though we should not be allowed to) then we get the series obove is equal to $\frac{1}{2}$. Now why is $\frac{1}{2}$ a sensible answer? If we look at the partial sums, we get the series, $$a_n=\frac{1}{2}(1+(-1)^n)$$. Now if we take the means of the partial sums, of $a_n$ we get $$\lim_{n\to\infty}\frac{1}{n+1}\Sigma_0^n a_n=\frac{1}{2}.$$ This is a compromise between the two points we are jumping between. However, this was not you question, but this should be a sort of warm up, so let us try to do something similar with your series and see if we can extract $\frac{1}{4}$.
We have the series $$1-2+3-4+5-6+\cdots.$$ If we  look at the partial sums as before, we get $$1,-1, 2, -2, 3, -3, 4,-4,\cdots.$$ Now let us take the means of the partial sums of this new sequence. This gives us $$1,0,\frac{2}{3},0,\frac{3}{5},0,\frac{5}{7},0\cdots.$$ Note that this is the result of two sub-sequences weaved together. The firts is the sub-sequence defined by $$\frac{n}{2n+1}$$, and the other is the constant zero sequence. The sequence $$\frac{n}{2n+1}$$ has limit $\frac{1}{2}$ while the constant zero sequence has limit zero. Now if we compromise between the two subsequences we get the desired $\frac{1}{4}$. This is related to Cesaro Summation. In fact this is just an iterated version.