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

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.

• The problem is those sequences are divergent. In general, you can make such sequences claim otherwise crazy things. Like, $1+2+3+4+5+\dots = -\frac{1}{12}$, which actually has some meaning in certain contexts... – mojambo Jan 18 '14 at 22:45
• @D.Clark: I don't know if I should be ashamed but I actually did do that. – Nick Jan 18 '14 at 22:53
• The thing about dealing with things that are undefined is that when you start treating them as if they are defined then weird things happen. – Clive Newstead Jan 18 '14 at 22:54
• @Clive: Why is it so? Are the results I get in doing this meaningless? What does the fact that I can derive a psuedo-sum tell me about the series? Doesn't the sum have some significant relationship with some aspect of the series? – Nick Jan 18 '14 at 23:01
• Rearrangements are far from innocent even if the sum is conditionally convergent: see Riemann rearrangement theorem – Ian Mateus Jan 19 '14 at 2:26

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.

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.

• I am aware that $\sum\limits_{n=1}^\infty n(-1)^{n-1}$ does not converge but by my manipulation, it's double converges and hence I get can get a psuedo-sum for it. Aren't I just giving a sum for it if it were to converge. Would I be wrong to do that? – Nick Jan 18 '14 at 22:51
• Unfortunately if it doesn't converge you can't multiply series by a number since the sum is not defined. I assumed that you define this sum in classical way $\sum_{n=0}^{\infty} a_n= \lim_{n \to \infty} S_n$ where $S_n = a_1 + \ldots + a_n$. If not please state that explicitly. – Konrad Szałwiński Jan 18 '14 at 23:03
• Doesn't the psuedo-sum have some significant relationship with any aspect of the series? What does $1/4$ mean to $\sum\limits_{n=1}^\infty n(-1)^{n-1}$ ? – Nick Jan 18 '14 at 23:13
• What is pseudo-sum for you and how you relate $1/4$ to $\sum_{n=0}^{\infty}$? The only way to sum this series up is to change the definition of summing. Sometimes we use Cesaro sum that is defined by $T_n = \frac{s_1+\ldots+s_n}{n}$ and we take a limit of $T_n$ instead of $S_n = a_1 + \ldots a_n$ but that has to be explicitly stated. – Konrad Szałwiński Jan 18 '14 at 23:56

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.

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).

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 $\lim_{x\to 1^-} f_3(x)=1/2$.