$2 - 4 + 6 - 8 + \cdots = 1 - 1 + 1 -1 + \cdots$?

Question

I want to talk about the weirdness of $2\sum\limits_{n=1}^\infty n(-1)^{n-1}$ ,

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

$$\times 2 \implies 2\sum\limits_{n=1}^\infty n(-1)^{n-1} = 2 - 4 + 6 - 8 + 10 - 12 + \dots\quad\longleftarrow (1)$$

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But

$2\sum\limits_{n=1}^\infty n(-1)^{n-1}$

$=\sum\limits_{n=1}^\infty n(-1)^{n-1} + \sum\limits_{n=1}^\infty n(-1)^{n-1}$

$= 1 - 2 + 3 - 4 + 5 - 6 + \dots$

$\quad\quad+ 1 - 2 + 3 - 4 + 5 - 6 +\dots$

$= 1 - 1 + 1 - 1 + 1 - 1 + 1 - \dots \quad\longleftarrow (2)$

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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.}

Answer 1

Just as others have pointed out, this is a false derivation, as the validity of every step is premised on the series being convergent. The very fact the series are divergent, gives you NON-UNIQUE and contradictory results just as you have demonstrated. The assignment of a finite number to a divergent series is but a mnemonic to denote the application of analytic continuation.

A typical example of this mnemonic representation would be the series representation of the Riemann zeta function $\zeta(s)$. $\zeta(s)$ is defined and a meromorphic function on the entire complex plane with a simple pole at $s=1$. The infinite series representation only truly makes sense and valid only for half of the complex plane $\mathcal R(s)>1$, even though the function $\zeta(s)$ as a meromorphic function is uniquely defined, or, if we use the correct nomenclature, uniquely analytically continued, through out the entire complex plane. For convenience (euphemism for laziness, which I do not quite understand, since one can simply use the name Riemann zeta function or symbol $\zeta(s)$ which is much shorter than writing out $\sum\limits_{k=1}^\infty \frac{1}{k^s}$)some people continue to use the series representation beyond its valid domain $\mathcal R(s)<1$. But that is just a mnemonic "symbol".

Just as others have suggested, Terry Tao has a great real analysis alternative to explain and treat the analytic continuation problem.

Answer 2

You are making a mistake at the very beginning. You assume that $\sum_{n=0}^{\infty} n(-1)^{n-1}$ exists. If fact it doens't as $n(-1)^{n-1}$ doesn't converge to $0$. These manipulations would be valid if had proven that this series converges. Please consult first theorem here.

Answer 3

The confusion and apparent contradictions lie in how we define infinite series. Indeed, $1-1+1-1+1-\cdots$, sometimes called Grandi's Series, can be evaluated to $\frac12$ using the geometric series test, but in order to apply that formula, we must have an appropriate notion of an infinite series.

To avoid these issues, we formally define infinite series as the limit of their partial sums. Since partial sums of $1-1+1\cdots$ likewise alternate as $1,0,1,0,1,0,\ldots$, by definition, the Grandi's series diverges.

The same caveat applies to your series.

Answer 4

As pointed out in the other answers, the series you manipulate do not converge (i.e. the sums do not have well defined values), so the manipulations are not valid.

However, there are some ways in which you are correct. For example, a lot of physicists would agree with you. See, for example, this very approachable exposition for a manipulation just like yours. Also, if you took an extension of the notion of a sum - for instance, Cesaro summation, you will find that the sums agree and your proof works (or can be modified to one that does).