Let's say I have to expand $(1+x)^{-1}$ using binomial expansion.

Using the theorem, I get:

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

Substituting $x$ for $1$, I get:

$$\frac{1}{2}= 1-1+1-1+1-1+1+....+{\infty}$$

A similar result arises with higher power of the exponent

For $(1+x)^{-2}$ we get:

$$(1+x)^{-2} = 1-2x+3x^2-4x^3+5x^4-6x^5+7x^6+....+{\infty}$$

Substituting $x$ for $1$, I get:

$$\frac{1}{4}= 1-2+3-4+5-6+7+....+{\infty}$$

How does this makes sense? Help please!

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    $\begingroup$ It doesn't. It only does if the series converge. $\endgroup$ – Bobson Dugnutt May 17 '17 at 10:54
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    $\begingroup$ It is interesting that this is the result that you get, as these appear to be the Cesàro summations for the relevant series. Perhaps someone can point out a deeper link? $\endgroup$ – Thomas Russell May 17 '17 at 10:55
  • $\begingroup$ @Lovsovs That's what I am asking. It should make sense right? Can someone explain how? $\endgroup$ – Chirag Arora May 17 '17 at 10:56
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    $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee May 17 '17 at 10:57
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    $\begingroup$ Before substituting any value in a power series, check if it lies within the radius of convergence. $\endgroup$ – StubbornAtom May 17 '17 at 10:58

Some of the most widely misunderstood ideas in mathematics are to do with divergent series and about when it's applicable to assign a value to a series (see all the nonsense on this site about $-1/12$ everything else).

The point is: certain equalities only hold in certain places.

Imagine you've got a number line. Sometimes it's only valid to do a particular operation in one part of the number-line and outside of this region, it doesn't make sense. Your question is an example of this.

And yes, it's true that $1-\frac12+\frac14-\frac18+\ldots=\frac{2}{1+2}=\frac23$ but this doesn't necessarily generalise * to $1-2+4-8+\ldots$. Clearly the first series gets closer to $0.666\ldots$ as you add more terms but the second series doesn't get closer to anything, it just keeps going to $\pm\infty$. So the first series is called convergent (since it approaches something) and the second is called divergent (since it doesn't).

To be explicit, the series $1-x+x^2-x^3+\ldots$ is only equal to anything ** when $x$ is between $-1$ and $1$ (not including the endpoints $1$ or $-1$). In essence, the value of $x$ needs to be small enough so that when you add more terms in the series, the terms get smaller and smaller until they're effectively $0$.

There needs to be a limit of the series for it to be equal to anything. i.e. there needs to be a finite value that the series gets closer to.

Does this make sense?

For completeness, there are alternative definitions of summation (Abel, Cesaro, analytic continuation) that do allow for summation in places where it's normally undefined. But this is not the normal definition.

* it doesn't generalise from $x=2$ to $x=-2$

** only equal to any finite number, using normal (not Abel) summation

  • $\begingroup$ No, it's not true that $1-\frac12+\frac14-\frac18+\ldots=\frac{1}{1+2}=\frac13$, it's $\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$, because $x=-\frac{1}{2}$ is well within the radius of convergence. $\endgroup$ – Professor Vector May 17 '17 at 11:20
  • $\begingroup$ @ProfessorVector Cheers, I did that in my head and hadn't yet checked it. Sloppy, I know. $\endgroup$ – Jam May 17 '17 at 11:22
  • $\begingroup$ It's still wrong, unless you correct the $\frac{1}{1+2}$, too. $\endgroup$ – Professor Vector May 17 '17 at 11:24
  • $\begingroup$ @ProfessorVector Thanks again $\endgroup$ – Jam May 17 '17 at 11:25

The equality $$\sum_{n=0}^\infty x^n = \frac{1}{x+1}$$

is only true if $|x<1|$, so you cannot substitute $x=1$ into it and expect the result to work.

  • $\begingroup$ Can you explain this? youtube.com/watch?v=PCu_BNNI5x4 Humans don't know numbers that well as much they think they do :P $\endgroup$ – Chirag Arora May 17 '17 at 11:12
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    $\begingroup$ @ChiragArora Yes, I can explain that. It's sloppy math leading to a false result. There are ways of getting the result, it's called Ramanujan summation. But that doesn't mean the sum of $1-1+1-1+\cdots$ is equal to $\frac12$. It just means there exists some way of assigning sensible values to infinite series such that we can assign the value $\frac12$ to $1-1+1-1+\cdots$. $\endgroup$ – 5xum May 17 '17 at 11:14
  • $\begingroup$ Okay agreed. But doesn't divergence means to produce an infinte sum? Where do you think is the sum approaching infinity in this case? 0,1, and 1/2 are all finite values. $\endgroup$ – Chirag Arora May 17 '17 at 11:16
  • $\begingroup$ @ChiragArora Look up "radius of convergence". $\endgroup$ – Bobson Dugnutt May 17 '17 at 11:16
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    $\begingroup$ @ChiragArora Divergence is not the same as the sum being infinite. Divergence simply means "not convergence". A sequence (or series) is divergent if and only if it is not convergent. And the sum $1-1+1-1+\cdots$ is not convergent, because the sequence of its partial sums (which is $1,0,1,0,1,0\cdots$) is not convergent (because it does not have a limit). $\endgroup$ – 5xum May 17 '17 at 11:18

As the other answers and comments point out, the problem here is "divergent" series. A divergent series is hard to pin down, because it breaks most of the usual rules of arithmetic, like associativity and commutativity of addition.
The series you wrote could be evaluated like this:

$$1 - 1 + 1 -1 +\cdots = (1-1) + (1-1) + (1-1) + \cdots = 0+0+0+\cdots = 0$$

or like this:

$$1 - 1 + 1 -1 +\cdots =1+(-1+1) + (-1+1) + \cdots = 1+0+0+\cdots =1.$$

The associative property says that we can group the additions as we please without changing the result. Yet, when we do it with this divergent series, the result does change. Or we could apply commutativity:

$$1 - 1 + 1 -1 +\cdots = 1 -1 +1+1 -1 +1+1+1 -1 + \cdots = 0+1+2+3+\cdots =\infty.$$

By rearranging and regrouping, we can make your sum add up to any integer we want. Which means it doesn't really have a value. Which in turn means that it doesn't mean anything when you set it equal to $1/2.$


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