Sum: $1-2+3-4+5-6+...$

Question

If we forget all the rules about infinte sums what am I doing wrong?

$$1-2+3-4+5-6+...=\sum_{n=1}^{\infty} n(-1)^{n+1}$$

(with Grandi's series)

$$1,1+(-2)=-1,1+(-2)+3=2,1+(-2)+3+(-4)=-2,...$$ we got a sequence: $1,-1,2,-2,3,-3,4,-4,5,-5,...$ but that turns out to nothing and if we look to the averanges of the partial sums:

$$1,\frac{1+(-2)}{2},\frac{1+(-2)+3}{3},\frac{1+(-2)+3+(-4)}{4},\frac{1+(-2)+3+(-4)+5}{5},...$$ we got: $1,-\frac{1}{2},\frac{2}{3},-\frac{1}{2},\frac{3}{5},...$

If we take: $1+\frac{2}{3}+\frac{3}{5}+...$ we got:

$$1+\frac{2}{3}+\frac{3}{5}+...=\sum_{n=1}^{\infty} \frac{n}{-1+2n}$$

and then we found:

$$\lim_{n\to \infty} \frac{n}{-1+2n}=\frac{1}{2}$$

can we say that:

$$1-2+3-4+5-6+...=\sum_{n=1}^{\infty} n(-1)^{n+1}=\frac{1}{2}$$

Answer 1

One can say: $1 - 2 + 3 - 4 + \dots = \sum (-1)^n n = \frac{1}{4}$ e.g. Wikipedia. These can be formalized using Borel (re)summation or Euler summation

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To understand Euler's idea let's do the simpler alternating sum first:

$$ S = (1 - 1 ) + (1 - 1) + \dots = 1 + (-1 + 1 )+ (-1 + 1)+ \dots = 1 -S$$

Therefore $2S = 1$ and $S = \frac{1}{2}$. Here we have used that the obvious-looking additive identity: $$\bbox[5px, border: 2px solid #6B6CED]{ (-1 + 1 )+ (-1 + 1)+ \dots = - \big[ (1 -1) + (1 -1) + \dots \big]}$$

The analogous story for the sequence $+1,-2,+3,-4,\dots$ requires more manipulation but the story is essentially the same. Here we group by three terms

$$ S = 1 -2 + 3 -4 + \dots = 1- (2 - 3+4 - 5 + \dots )$$

What next? Split this arithmetic sequence into two terms:

$$ 1- (2 - 3+4 - 5 + \dots ) = 1 - \big[ 1-1+1-1+\dots \big] - \big[ 1 -2 + 3 -4 + \dots\big] = 1 - \frac{1}{2} - S $$

This gives you $\bbox[5px, border: 2px solid #6B6CED]{S = \frac{1}{4}}$

Beware plausible-looking manipulations can give you other numbers. While writing this I also showed $S = \color{red}{-}\frac{1}{4}$.

Answer 2

No, the series is divergent. You can't just "forget" the rules of mathematics. If you "forget" the rules, then you are not doing math. It doesn't matter that you would like the series to be convergent, it is a question about whether or not it (objectively) is.

About your series: You might remember the Test for Divergence. This rule says that if $$ \lim_{n\to \infty} a_n \neq 0 $$ (or if the series does not exist) then the series $$ \sum_{n=1}^{\infty} a_n $$ is divergent.

It, hopefully, is clear that for your sequence, where $a_n = n(-1)^{n+1}$ the limit $\lim_{n\to\infty} a_n$ is not zero. So the series is divergent.

Answer 3

Hint: What is the series of $\dfrac1{1+x}$ ? What happens when we multiply both sides with x, and then differentiate them ?

Answer 4

At least in contemporary English, the colloquial connotations of "rule" may be unhelpful here. That is, a "rule" is something imposed by an authority, and cannot be analyzed for sensibility or possibly even for intention or purpose. Perhaps in mathematics we can do better than invoke some anonymous (or not) authority about legitimacy, and, instead, talk about interpretation, meaning, intentions, and such.

In particular, an "infinite sum" requires interpretation. The current standard is that it means the limit (if it exists) of the (finite) partial sums. We do have an unequivocal sense of adding finitely-many things, and limits are arguably less ambiguous than infinite sums. By this standard, one can prove Cauchy's criterion, that if that limit of partial sums exists, then it is necessary that the summands ("terms") in the infinite sum must go to $0$ (although that is not sufficient for convergence). So, by this standard, the summands $n\cdot (-1)^{n+1}$ do not go to $0$, so the whole thing has no chance of converging.

But that is not the only question that can be asked, since there are other options ("summation methods") for obtaining "sums" of infinite series that may not converge by the "strict" standard. This is the theory of "divergent series", which is more than just declaring such a thing nonsensical or divergent per se. The more intelligible of these involve analytic continuation, whether explicitly or implicitly, and go back to Euler, at least, who considered "values" of the zeta function $\zeta(s)=\sum_{n\ge 1} 1/n^s$ at $s=-1,-2, \ldots$. The question here could be viewed in that context, if desired. More elementary analytic continuations bear upon things like $1-2+4-8+16-\ldots$ which is $1-x+x^2-x^3+\ldots=1/(1-x)$ at $x=2$, outside the region where the series converges in a strict sense, but the expression $1/(1-x)$ giving the analytic continuation certainly makes sense there, so we might declare $1-2+4-8+...=-1$.

Edit: in response to @Thomas' comment/question, and as in @Lucian's answer, a relatively simple way to appraise the given series is to use the geometric series $1/(1+x)=1-x+x^2-x^3+\ldots$, take its derivative $-1/(1+x)^2=-1+2x-3x^2+\ldots$, and "evaluate" at $x=1$. This gives $1/4$, as in @fibonatic's answer. To my mind, the real point here is seeing what the process produces.

More as Euler might have done, rather than using power series to analytically continue, use Dirichlet series $f(s)=\sum_{n\ge 1} (-1)^{n+1}/n^s$. This is close to "the" (Euler-Riemann) zeta function $\zeta(s)=\sum_{n\ge 1} 1/n^s$: $f(s)+2\cdot {1\over 2^s}\cdot \zeta(s)=\zeta(s)$, so $f(s)=\zeta(s)\cdot (1-2^{1-s})$. Various means to analytically continue $\zeta(s)$ give the (by now popularly notorious) summation $1+2+3+4+\ldots=\zeta(-1)=-1/12$, which was already "known" to Euler long ago, and to Ramanujan by his own idiosyncratic devices. Thus, the desired $f(-1)=\zeta(-1)\cdot (1-2^2)={-1\over 12}\cdot (-3)={1\over 4}$ again.

(Euler also "found" $1^2+2^2+3^2+\ldots=0$ and similarly for every even, positive power, which is consistent with the analytically continued $\zeta(s)$ values.)

So my reaction to the original question would be that the two most obvious summation devices don't give $1/2$, but $1/4$, ...

Answer 5

Lets call your summation series $S$,

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

If we use the Grandi's series, call it $G$, and agree on that its summation is equal to $\frac{1}{2}$, then by subtracting it from $S$ you get,

$$ S-G=(1-2+3-4+5-6+\dots)-(1-1+1-1+1-1+\dots)=(1-1)-(2-1)+(3-1)-(4-1)+(5-1)-(6-1)+\dots=-(1-2+3-4+5-\dots)=-S. $$

Solving this equation for $S$ yields,

$$ S=\frac{G}{2}=\frac{1}{4}. $$

Answer 6

Although the sum does not converge in the traditional sense, there are various regularization techniques that use analytic continuation to assign a value to these (and they are very important for renormalization of divergences in physics, especially string theory). Moreover, different regularization techniques usually produce the same value.

Start by finding a generating function (a series that for some value of $x$ has the same terms as your sum).

Let's start with $$1-1+1-1+\cdots = \sum (-1)^n = \sum (-1)^n x^n|_{x=1}=\frac{1}{1+x}|_{x=1}=\frac12$$ (evaluated at the boundary of the convergence radius so it's only true in the regularization sense).

If you take the derivative over $x$, you get

$$\sum (-1)^n n x^{n-1}=-\frac{1}{(1+x)^2}$$ Put $x=1$ in it: $$-1+2-3+4-5+6-\cdots = -\frac{1}{4}$$ Multiply by $-1$: $$1-2+3\cdots = \frac14$$

You should read this: https://en.wikipedia.org/wiki/Zeta_function_regularization

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Another, even more "suspicious" looking derivation is the following:

$$1-2+3-4+\cdots = $$ $$\begin{array}+1 & -1 & +1 & -1 & +1 & \cdots \\ & -1 & +1 & -1 & +1 & \cdots \\ & & +1 & -1 & +1 & \cdots \\ & & & -1 & +1 & \cdots \\ & & & & \vdots &\ddots \end{array}$$ $$=1\cdot (1-1+1\cdots)-1\cdot(1-1+1\cdots)+\cdots $$ $$=(1-1+1-1+1-1\cdots)^2=\frac14$$

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To actually answer your question: what's wrong is that you for got about the $-1/2$ terms in the average partial sums. The sequence of average partial sums does not converge (of course not), it alternates in sign (limiting cycle $\pm 1/2$) so the limiting step of the odd terms doesn't tell you anything.

You aren't even allowed to reorder terms in conditionally convergent sums (you could get any result, even infinity), and divergent sums are even worse. You will notice that all the solutions you can read here keep the terms in order without skipping anything. Just recall what happens if you exchange neighbours in the series $1-1+1-1+1-1\cdots$ - it actually flips the sign!

If you want to evaluate divergent sums, you will need some regularization step: either analytic continuation or an implicit step to replace iteration with equation solving (such as expressing the series with itself) and you will need to do it without reordering the series.

Answer 7

There is some valuable commentary in the other answers pertaining to this series, but no one has addressed your specific question! It looks like you were trying to compute the Hölder sum of the series. A Hölder sum operates by repeatedly averaging the sequence of partial sums. For this particular series, the first Hölder sum does not exist; only the second and higher sums are $1/4$.

Let's go step by step. I need some notation:

• Let $t = (1,-2,3,-4,5,-6,\ldots)$ be the sequence of terms of the series that we're most interested in.
• For a sequence $x$, let $S(x)=(x_1,x_1+x_2,x_1+x_2+x_3,\ldots)$ be the sequence of partial sums.
• For a sequence $x$, let $D(x)=(x_1,x_2/2,x_3/3,\ldots,x_n/n,\ldots)$ be the sequence formed by dividing each term by its index.
• For a sequence $x$, let $M(x)=D(S(x))$ be the sequence of means of $x$.

First, you observed that the partial sums of the series are:

$S(t) = (1,-1,2,-2,3,-3,4,-4,5,-5,\ldots)$

Then you computed $M(t)=D(S(t)) = (1,-\frac{1}{2},\frac{2}{3},-\frac{1}{2},\frac{3}{5},\ldots)$. However, that isn't the relevant sequence of averages. We actually want $M(S(t))=D(S(S(t)))$. First we compute

$S(S(t)) = (1,0,2,0,3,0,4,0,\ldots)$

and then

$D(S(S(t))) = (1,0,\frac23,0,\frac35,0,\frac47,0,\ldots)$.

Now we can see that $M(S(t)) = D(S(S(t)))$ is also a divergent series, but we've made some progress. Let's apply $M$ again:

$S(D(S(S(t)))) = (1,1,\frac53,\frac53,\frac{47}{21},\frac{47}{21}\ldots)$

and

$D(S(D(S(S(t))))) = (1,\frac12,\frac59,\frac{5}{12},\frac{47}{105},\frac{47}{126},\ldots)$.

At this point it should be plausible that $\lim D(S(D(S(S(t)))))=\frac14$. Of course, there are easier ways to arrive at $\frac14$. My point is to show how your strategy is supposed to work. The short answer to where you went wrong is simply that you skipped an $S$!